o > > IT! W. A. J. CHAPMAN Senior Workshop Calculations Third Edition SI Units O 33 X o O > o c r 5 0) o r o r Senior workshop calculations l\1 b Senior workshop calculations Dr. W. A. J. Chapman MSc(Eng), FIMechE, HonFIProdE <D EDWARD ARNOLD ©W.A.J. Chapman 1972 First published 1941 by Edward Arnold (Publishers) Limited 25 Hill Street London WIX 8LL Reprinted 1944, 1947, 1949, 1952 Second edition 1954 Reprinted 1957, 1959, 1961, 1962, 1965 Third edition 1972 Reprinted 1973, 1978 ISBN 7131 3260 4 All Rights Reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or other- wise, without the prior permission of Edward Arnold (Publishers) Ltd Photoset and printed in Malta by Interprint (Malta) Ltd This book is a revision of the workshop practice material in Senior Workshop Calculations . This revision was made necessary by the advent of the SI System and of developments in some aspects of teaching and industrial practice. Senior Workshop Calculations has enjoyed a vogue of thirty years and is used in many parts of the world. From the various expressions I have received I am sure that it has rendered a useful contribution to the work of countless students as well as to that of mature workers engaged in engineering. During the revision it was decided to separate out the text dealing with practice calculations as a first priority. The text includes most of the material required by students and practitioners in workshop and production engineering practice and should provide a useful textbook for National Certificates and Diplomas, and for City & Guilds courses in Mechanical and Production Engineering subjects. In the work of revision I have received considerable advice and help from Mr M. G. Page, BSc(Eng), FIMechE, FIProdE, and I should like here to acknowledge my warm appreciation of his generous and kind assistance. I hope that in its present form the book will continue to serve the interests of students and other workers in those aspects of engineering it has always been my desire to foster. Hatfield, 1972 W. A. J. C. Contents 1 Introduction: The SI system of units and conventions Basic SI units - Supplementary and derived units - Multiples and sub-divisions of the unit - Representation of unit symbols and quantities - Dimensioning - Mass, weight and force - The kilogramme weight - Equations of motion - Acceleration 1 2 Measurements and gauging Limits and tolerances - Limit systems - Calculations of limits - Slip gauges - The spirit level - The sine bar - Gauging large radii and holes - Location of points on angular surfaces - Measurement of tapers with balls and rollers - 3-wire measurement of screw threads - Miscellaneous measurement problems 13 3 Calculations for cutting, turning and boring Speeds and feeds - Arithmetic and geometric progression - Cutting tool life - Tool angles - Effect of tool height - Taper turning - True shape of form tools, flat and circular - Approximate change wheels by continued fractions - Cutting power for turning and drilling 56 4 Calculations for gears and gearcutting Involute gear teeth - The tooth Vernier - Constant chord measurement - Module pitch - Plug Method of checking gear teeth - Base pitch - Stub teeth - Backlash - Helical (spiral) gear calculations - Worm gearing - Bevel gearing 84 5 Milling and the milling machine Milling cutters - Cutter teeth - Rake - Clearance - Tooth angle - Tooth grinding - Helical teeth - Speeds and feeds for milling cutters - Milling power calculations - The dividing head - Simple, compound, differential, angular indexing - Spiral milling - Cam milling - Solid angles 113 6 Mechanical principles I Vectorial representation - Addition and subtraction of vectors - Application of vectors- forces on a cutting tool - Balancing of faceplates - Vector velocity diagrams for mechanisms - Moment of a force - Parallel forces 148 7 Mechanical principles II Friction - Machines and efficiency - The inclined plane and screw - Bearings - Bearing pressure - Bearing friction - Stress and strain 176 8 Mechanical principles III Velocity - Acceleration - Equations of motion - Force - Energy; kinetic and potential - Circular motion - Accelerating torque - Energy of flywheels - Temperature - Heat - Amount of heat - Heat energy 198 Appendices I and II: ISO standard holes and shafts 225-6 Appendices III and IV: BSI standard holes and shafts 226-7 Appendix V: Conversion table 228-9 Appendix VI: The trigonometrical addition formulae 230 Appendix VII : Continued fractions 23 1 Answers 235 Index 242 1 Introduction The SI system of units and conventions The initials SI are an abbreviation for Systeme International d'Unites (International System of Units), the modern form of the metric system, finally agreed upon at an international conference in 1960. It is now being adopted widely throughout the world and is likely to become the primary world system for units and measurement. As we shall discuss below, the system rationalises the main metric units of measurement and standardises their names and symbolic representation. It also rationalises certain mechanical principles and conventions. The British system of weights and measures is many centuries old, and the derivation of its units with their multiples and sub-divisions is often obscure. The system has been refurbished from time to time but the yard and the pound with their multiples (e.g. mile) and sub-divisions (e.g. ounce) have persisted; so have such ridiculous measures as 5| yards = 1 rod, pole or perch, 4 roods = 1 acre or 141b = 1 stone remained with us to try the mental agility of generations of students, not to mention the more mature, and less mentally agile population. The metric system was founded during the French Revolution and has been adopted for use by most countries with the notable exceptions of the British Commonwealth and the U.S.A., but even in these countries it is used for precise scientific measurements. The basic units of the SI metric system are the metre and the kilo- gramme and it is exclusively decimal, so that all multiples and sub- divisions of the standard are found by applying factors of 10 (1 kilometre = 1000 metres; 1 kilogramme = 1000 grammes; 1 hectare (area) = 100 x 100 square metres, and so on) . In the English system, of course, there is no such orderly pattern and indeed, the foreigner might well question our sanity when he hears us refer to 1 12 pounds as a hundredweight. However, we have, at last, been caught up with the progress of the times and as a nation we have decided to change over to the metric system. The entire text of this book conforms with the SI system and the object of this introductory chapter is to provide help and reference for the reader as he finds his way into what may seem, at first, to be a complexity. The best 2 INTRODUCTION advice that can be given for achieving rapid progress in coping with the change is to become familiar with the new measures and to learn, as soon as possible to think in terms of them, and not to persist in making mental conversions back to the old units. This process can be speeded up by acquiring, as soon as possible, a mental appreciation of a range of lengths, weights, capacities, etc. Some of these could be (say) the mental judge- ment of 25 millimetres, 1 metre, the weight of 1 kilogramme, the amount of fluid comprising 1 litre (1000 cm 3 ) and the pressure corresponding to 1 bar (approx 1 atmosphere)* . In this way it will soon be possible to think of these measures in their own right and not grope around converting them to their English equivalents (1kg = 2-21b; 1 litre = about If pints and so on). A similar process is concerned in the learning of a foreign language where fluency will never be achieved until a student thinks in terms of the language concerned and abandons all attempts to interpret mentally from, and into, English. It is well known that another language is quickest learned by living amongst those who speak nothing else. If the reader can approach this new mathematical and scientific language in this frame of mind he will find that the former system will rapidly re- cede, so increasing the ease with which he can cope with the problems involved. Basic SI units The SI system is based on six primary units as follows: Table 1. Basic SI units Quantity Unit Symbol Length Metre m Mass Kilogramme kg Time Second s Electric current Ampere A Temperature Kelvin K Luminous Intensity Candela cd In addition to these there are a number of supplementary and derived units. We give below a selection of those which are most likely to be required by students using this book. A full list of the SI Units and con- ventions is given in BSI publication PD 5686. *The reader has probably heard of the bar in reference to millibars of atmospheric pressure in Met Office forecasts. SUPPLEMENTARY AND DERIVED SI UNITS Supplementary and derived units Quantity Table 2. Selected supplementary and derived SI units Unit Symbol Area Square metre m 2 Volume Cubic metre m 3 Density Kilogramme per cubic metre kg/m 3 Velocity Metre per second m/s Acceleration Metre per second squared m/s 2 Force Newton N(kgm/s 2 ) Moment of force Newton metre Nm Pressure, stress Newton per square metre N/m 2 Work, energy Joule J(Nm) Heat quantity Power Watt W (J/s) Plane angle Radian rad Temperature (Everyday use) Degree Celsius* °C Specific heat capacity Joule per kilogramme degree Celsius J/kg°C Electric tension ) Potential difference > Volt V Electromotive force ) *It is probable that the word "centigrade" will remain, but SI recommends the use of Celsius to prevent confusion with another unit. Multiples and sub-divisions of the unit In the same way, that in the British system where the yard is divided into feet and inches and multiplied into the furlong and mile, the pound multi- plied into the cwt and ton and divided into the ounce, so it is necessary to make similar provisions in the SI system. One of the advantages of the system for this purpose is the simplicity of its multiples and sub-multiples because of its decimal (10) character. We will give, again, a selection of the chief factors likely to be required by the reader leaving him to study the BSI literature if he wishes to pursue the remainder. It will be noticed that multiples and divisions involving 10 4 , 10 5 and certain higher powers are not included in the system. This has been done in order to rationalise the procedure by using, after 10 2 and 10~ 2 , only powers which are multiples of 3, and the full list of such factors extends from 10 12 at the higher end to 10~ 18 at the lower. INTRODUCTION Table 3. SI multiples and sub- multiples Factor by which unit is multiplied Prefix Symbol Example One million = 10 6 mega M megawatt (MW) One thousand = 10 3 kilo k kilometre (km) One hundred = 10 2 hecto h hectare (ha) Ten = 10 deca da decagram (dag) One tenth = 10-' deci d decimetre (dm) One hundredth = 10 ~ 2 centi c centimetre (cm) One thousandth = 10 ~ 3 milli m millilitre (ml) One millionth = 10 - 6 micro /" microvolt (juV) Having now considered the quantitative manipulation of the basic units we may now go on to consider the facilities available for alternative smaller divisions, or larger multiples of the main units, and of certain additional variations allowable in the system. By this means it is possible to employ a unit of suitable proportions for any particular set of circum- stances. (Table 4). Representation of unit symbols and quantities It often happens that when we move into a new house, or office, or take on something which changes our way of life we take the opportunity of overhauling our methods. So it is with the changeover to this new system where the adoption of SI is accompanied by various conventions regarding the presentation of information. These are summarised as follows: (a) Writing the unit symbols (i) The full stop, usually placed after an abbreviated word, is never used after an abbreviated symbol except at the end of a sentence. Thus, 50kg, 10m, 8 kg, 30s, 5N/mm 2 , all without a full stop, but: "the vehicle had a mass of 1100 kg." end of sentence, full stop. (ii) The plural V is not used. (50cm not 50 cms) (iii) The proposition "per" is replaced by the oblique / (rev/min not rev per min) (iv) Symbols for areas and volumes are qualified by the index, (cm 2 not sq cm and cm 3 not cu cm) (Note that when converting quantities denoted by indices: lm 3 = (100cm) 3 = lOCFcm 3 = 10 6 cm 3 ) (continued on p. 6). REPRESENTATION OF UNIT SYMBOLS AND QUANTITIES Table 4. Recommended multiples and sub-divisions of the basic SI units Quantity Plane angle Length Area Volume Time Velocity Mass Other units acceptable Multiple or in the system and sub-multiple likely to persist m rad degree minute' second" km cm* mm /urn micron (i4mm) km 2 hectare (ha)(10 4 m 2 ) cm 2 are (a)(10 2 m 2 ) mm 2 dm 3 litre (/)(100 cm 3 ) cm 3 millilitre (ml) mm 3 Ms year, month, week. ks day (d) hour (h) ms minute (min) /US Mg km/h tonne (OR metric ton) (t) metric carat (2 x 10 - 4 kg) mg Density kg/ dm 3 or kg/1 g/cm 3 g/1 Force MN : kN : mN : pti kgf (weight of 1 kg mass) (not included in SI system but likely to be used) 1 kgf = 9-806N Moment of force MNm:kNm:,uNm Pressure N/mm 2 : N/cm 2 : kN/mm 2 bar = 10N/cm 2 mN/m 2 :/xN/m 2 hectobar (hbar) = 10 3 N/cm 2 (Will probably be used for Stress kN/mm 2 : kN/cm 2 : N/mm 2 pneumatic pressures. High N/cm 2 :kN/m 2 pressures and stresses will be expressed in units shown opposite) Work and energy MJ:kJ:mJ Kilowatt hour (kWh = 3-6 MJ) Power MW:kW:mW:/iW Specific heat capacity kJ/kg°C * The cm is not recommended for general use and it is hoped that it will eventually disappear. It is, however, a very convenient unit for certain purposes as will be seen from the uses of it later in our text. It will be observed, however, that the cm 2 and cm 3 are still permissible. 6 INTRODUCTION (b) Numerical values (i) When a quantity is less than unity ( 1 ) always place zero (0) before the decimal point. (0-625:0.0031, etc.) (ii) As far as possible always express a quantity in terms of a power of 10, so using the index of 10 instead of a row of 0's. (36-2 x 10 3 instead of 36 200; 1-5 x 10~ 3 instead of 0-0015) and preferably use powers of 10 which are multiples of 3. (hi) Separate a row of digits into groups of three by a space, instead of using a comma. (71 562 instead of 71,562, or 0-006 13 instead of 0-006, 1 3) . But a group of four digits may be left without separa- tion (e.g. 6713 or 0-0036 without separation). Dimensioning (drawings) It is customary, when dimensioning drawings, to dimension all sizes in millimetres, and not to write the unit (mm) after the dimension. An instruction may be given to the effect that all sizes are in millimetres but this is not always done. All the diagrams in this book are dimensioned according to this rule so that the reader will now recognise that <^^ > means 65 mm. Mass, weight and force The reader will observe that kg is the SI unit for MASS and that a unit for weight is not mentioned in the scheme. There is, however, the Newton (N) as the unit of FORCE. The mass of a body is the amount of matter (or material) of which it is composed and the kilogramme unit is equal to the mass of the international prototype kilogramme which is in the custody of the Bureau International des Poids et Mesures at Sevres near Paris. The only means of measuring and comparing masses is by weighing them so that the weight of an object is proportional to its mass and indeed, the weight of a body is the downward force its mass exerts under the influence of the earth's gravitational pull. We have seen this concept very effectively illustrated since astronauts have been penetrating beyond the influence of the gravitational pull of the earth and most readers will have seen the fascinating television pictures of the interior of the capsule, with objects floating about, still having their mass, but without weight. The definition of a newton (N) of force is that force which acting on one kilogramme of mass will propel it along with an acceleration of one metre per second per second (i.e. it gains 1 m per second every second). Put into symbols this becomes: IN = lkgm/s 2 THE KILOGRAMME WEIGHT (kgf) 7 The kilogramme weight (kgf) The gravitational pull where we are (England) is such that it imparts an acceleration of about 9-807 metres per second per second (9-807 m/s 2 ) on a freely falling body and this pull, acting on a mass of 1 kg causes it to exert a downwards force (its weight) of 9-807 Newton of force (since 1 Newton = 1 kgm/s 2 ). Thus the weight of a mass of 1 kg expressed in newtons of force is: lkg weight (kgf) = 9-807 newtons (usually approximated to 9-81 N) The weight of a body having a mass of 150 kg = 150 x 9-81 = 1471-5 N) The mass of a body is a quantity which never changes (unless a piece is cut off or added to it), but the force of gravity varies slightly on dif- ferent parts of the earth. This means that the weight of 1 kg of mass in England will not be quite the same as its weight at the equator. The variation is very small (about 0-5%) over the surface of the earth so that for all but the very accurate scientific work there is a justification for using the kgf unit of weight in everyday life. The reader will find kg loosely referred to as the "weight" of an article, when what is really meant is the gravitational pull on a kg of mass. However, for some problems, such as those involving weights and costs of materials, it will be convenient to work in kgf rather than newtons, since suppliers of materials will always quote kg or tonne, and not newtons as the weight in their price lists and specifications. The following examples will illustrate the use of force units: Example 1. A round steel bar, 100 mm diameter, 1 metre long is placed standing on its end. If the contact between the end of the bar and its support is uniformly spread over the whole end face of the bar calculate the intensity of pressure over the area of contact. Take the density of steel as 7-83 g/cm 3 . Mass of bar = (Volume)(density) = j X 10 2 x 100 x 7-83 grams (Working in cm) 0-7854 x 10 4 x 7-83. = __ kg = 61-5 kg Downward force (weight) exerted by bar = (mass) (gravity) === 61-5 X 9-807 = 603 N Area of bar end = 0-7854 x 10 2 cm 2 = 78-54 cm 2 INTRODUCTION T . c force 603 Intensity of pressure = area 78-54 = 7-68 N/cm 2 Example 2. A vehicle having a mass (m) of 2000 kg has a fractional resistance to motion of m/20. If the drive is equivalent to an average force of 320 N find the speed attained after 1 min from rest. 2000 Resistance to motion = -^— N = 100 N Effective driving force = 320N - 100N = 220N and since force (F) = mass (m) x accel (a) 220N = 2000 a 220 All ,2 " = 2000 =°- llm/s2 Speed after 1 min V = (accel)(time) = 0-11 X 60 = 6-6 m/s Example 3. A pin is being driven home by a hammer of mass 1 kg. When the hammer is moving at 1-5 m/s it strikes the pin and drives it 10 mm further in. Assuming the hammer is brought to rest at constant decelera- tion estimate the average force of the blow. If the hammer travels 10mm (0-01 m) whilst decelerating at a constant rate from 1-5 m/s to rest: Average speed over the stopping period = ¥- = 0-75 m/s and the space moved = 0-01 m Hence: duration of the blow (t) = ~-=j — 0-0133 s We then have for the movement of the hammer after it strikes the pin: initial speed (u) = 1-5 m/s: final speed (v) = and time (t) = 0-0133 and from v = u + at (see p. 10) = 1-5 + 0-0133a EQUATIONS OF MOTION from which a = nnn o = - 112-7 m/s 2 (retardation) The average force of the blow is found from: F = (mass)(accel) = 1 x 112-7 = 112-7N The equations of motion The reader, no doubt, will have already realised that the SI system is more orderly and coherent than the British system and this might be pursued by considering its application to the equations of motion. If s = space travelled, v = velocity (or speed) and t = time then s = vt or v = - (1) The basic unit for s is the metre (m) and for t, the second. This gives us g the secondary basic unit for velocity as v = - = m/s (metres per second). Naturally, it is not always desirable, or possible, to work in metres and seconds but alternative units of larger or smaller dimension are avail- able to suit the conditions or aspects of any particular situation, (e.g. for a road speed we should probably use kilometres and hours) . Thus: 1 kilometre = 1000 metres 1 hour = 3600 seconds 1 km 1000 m _ 5 m _ 5 1 h ~ 3600 s ~ 18 s ~ 18 m/S i.e. to convert km/h to m/s multiply by -j-~. Acceleration Acceleration is the rate of change of velocity , . change in velocity i.e. acceleration = time taken for the change If a body starts from rest and acquires a velocity of v after time t, and using the symbol a for acceleration v a = — t KJ INTRODUCTION Since the basic derived unit for v is m/s, and the basic unit for t is s, so the basic unit for acceleration is— — 5- s = — (metres per second squared). Transposing the above a = - to give v we get v = at. If, instead of starting from rest, the body already had an initial velocity of u. Then, final velocity after time t: f = u + at (2) Instead of accelerating a body may be slowing down or decelerating. Then the acceleration will be a minus quantity and u — at This slowing down is termed a retardation. The above relationships may be illustrated graphically and this is shown at Fig. 1 . Fig. 1 Example 4. A drop stamp falls freely for 5 metres under the action of gravity. Find: (a) the time of fall and (b) its velocity at the instant it strikes the tup. Takeg = 9-81 m/s 2 . ACCELERATION 1 1 The graph representing the fall is shown at Fig. 2 and if v is the final velocity after time t: Vel. O t u \ArM\ 5> s^. 5m\ " Time Fig. 2 We get: 1 , C A 10 5 vt = 5 and v = — t But for accelerated motion from rest v = at where a = g = 9-81 m/s 2 Hence, equating this to (1): 10 = a = 9-81 t and t 2 10 9-81 Substituting in (1) for t we have: v — 10 1-01 t *= 1-01 s = 9-9 m/s (1) Example 5. A machine ram is operating on a stroke of 360 mm. It starts from rest, accelerates at a uniform rate until the centre of the stroke and then retards at a uniform rate to a standstill at the end of the stroke. If the stroke occupies 1-5 seconds find the acceleration and the maximum speed attained. 12 INTRODUCTION This problem is best solved with the help of a graph and the motion is represented in Fig. 3. Time Fig. 3 The ram accelerates from to A, its velocity increasing uniformly, and the reverse process takes place from A to B. The area OAB represents the space travelled which, in this case, is 360 mm (0.36m). Hence 0-36 m = KOB)(AC), and since OB = l-5s 0-36m = |(l-5s)(AC) 2 x 0-36 m l-5s AC = v = For half the stroke v = at v 048 Hence a = - = T^^m/s = 0-64m/ s 2 = 0-48 m/s 2 Measurement and gauging Limits and tolerances When parts which must fit together are being made under conditions which do not permit of each fitting pair to be mated up individually, it becomes necessary to arrange the working dimensions so that if each component is made to them, the required type of fit will be assured. To achieve this, a system of limits and fits is adopted. The definitions used in connection with limit systems will be gathered from the following and Fig. 4. , jTolerance '^///(//s r V ' __ o 1 1 I 1 oo -S J 1 ~i — "^ so -J 00 - — i S9 "^ *« °> -c os On .c o A ^ - .2>^ ^- .o> fN ra "> a: '. K> -J 1 ** a: K) i ' w/z/Jm Sha it 519 n 31-9 80 t 68 i 1> Holt 32 ! 31 9 15 A 85? Fig. 4 The low limit is the dimension of the smallest permissible size, and the high limit the largest permissible size. The component is acceptable if its size lies anywhere between the two. The tolerance is the difference between the two limiting dimensions. [The tolerance on the above shaft is 31-980 - 31-968 = 0.012mm.] The allowance is the variation between the sizes of the hole and shaft necessary to give the type of fit required. For a running fit the shaft must be smaller than the hole (clearance fit), whilst for a driving or force fit, the shaft must be larger than the hole (interference fit). Between the extremes of clearance and interference there is a range of fits such as push fit, slide fit, etc., in which there is only a small variation between the hole and shaft sizes (transition fit). 14 MEASUREMENT AND GAUGING Limit systems The numerical value of the limits for any related hole and shaft will depend on: (a) The nominal size (i.e. whether 25mm, 50mm, 100mm, etc.) (b) The class of fit (e,g. running fit, push fit, force fit, etc.) (c) The grade of workmanship desired. The ISO system, set out in BS Specification No. 4500 (1969), allows for 27 types of fit and 18 grades of tolerance over a size range of zero to 3150mm. At first sight this seems an enormous provision, but the fits and grades of workmanship covered allow for everything from fine gauge work to the roughest form of production, and even for some classes of raw materials. Average workshop requirements may be met from a limited part of the specification and for this purpose suitable recom- mendations are given. In the system the 27 possible holes are designated by capital letters A B C D etc., and the shafts by small letters covering the same range. The 18 accuracy grades are denoted by the numbers 01, 0, 1, 2, ... 16. The nomenclature adopted for specifying any particular hole or shaft with its tolerance grade is to write the hole or shaft letter with its grade number: thus H7 for a hole, or e8 for a shaft of the fit and accuracy given by the letter and numeral concerned. A fit involving these two elements is written H7-e8 or H7/e8 . SELECTED FITS (HOLE BASIS) Type of Fit Clearance (Slack, Running etc.) Transition (Push, Slide etc.) Interference (force, Driveetc.) (Details of the limits for the above shafts over a diameter range 6 mm to 250mm are given in Appendix 1, page 225.) LIMIT SYSTEMS 15 For average workshop use the H hole associated with the accuracy grades 7 to 11 (H7 to Hll) are recommended as being satisfactory and details of the limits of these for a diameter range of 6mm to 250mm are shown in the table on p. 226. 140 120 100 80 60 40 20 E ° E S 20 o £ 60 v. ® "o 80 100 120 140 160 180 200 220 240 - ill Clearance Transition Interfere/! ce Fig. 5 ISO System of Limits and Fits. Hole and shaft relationships for selected fits (hole basis). (Tolerance scale applies to the diameter range: 18mm— 30mm.) 16 MEASUREMENT AND GAUGING From the shafts included in the specification the table gives a selec- tion of those recommended as likely to be the most useful in coping with the needs of the average workshop. The holes with which they should be associated as well as the approximate type of fit are also given. Naturally from a selection of 27 fits associated with 18 grades of ac- curacy it is possible to choose many combinations, but the small selec- tion above should be sufficient to help the reader in his study of the subject. Fig. 5 illustrates the hole and shaft relationship for a selection of the fits associated with the H7 to Hll holes as recommended in the BS Specification. From the diagram the reader will be able to trace the maximum and minimum metal conditions for the range of diameters to which the details apply. Exercises 2a Write down for problems Nos. 1 to 4 the hole and shaft limits, and calculate the maxi- mum and minimum clearance or interference. (See tables in appendices) 1. 75mm BSI H10 hole and e8 shaft. 2. 35 mm BSI H9 hole and k6 shaft. 3. 20mm BSI H7 hole and f7 shaft. 4. 57 mm BSI H8 hole and s7 shaft. 5. Two 32mm BSI H8 holes are bored in a plate at 105 mm ±0-02 mm centres. Their centre distance is to be checked by a gauge of the type shown at Fig. 6. If the "go" ends Fig. 6 of the checking plugs are 12 mm diameter calculate their "Not go" diameters and their centre distance. 6. To allow for gauge wear the "Go" end of a limit plug gauge, made for a 40 mm BSI H9 hole, is made larger than the minimum hole size by 10% of the tolerance. Calculate the diameter of this end of the gauge. 7. A limit caliper gauge made for a 50 mm BSI f7 shaft is to be ground out to suit a k7 shaft. Determine the alteration necessary. 8. Two blocks each 25 mm wide are made to BSI h7 limits and a 50 mm slot is made to H8 limits. Compare the fit of these blocks put together with that of a single 50 mm block SLIP GAUGES 17 made to h7 limits. (Assume the width of the blocks put together to be equal to the sum of their individual thicknesses.) 9. A number of limit plug gauges are available, made for the former Newall fin. Class B hole ( — in). Could these be ground down to suit a 19mm BSI H8 hole and, if so, what alteration would be necessary? 10. A BS 40mm b9 shaft is placed in an H8 hole. Determine the greatest and least clearance possible between the shaft and the sides of the hole. Slip gauges For the purpose of checking the accuracy of micrometers, verniers and other gauges, it is necessary to have available some means of building up any required length. In most workshops this is achieved by the use of slip gauges. These consist of blocks of different thicknesses, which are made to such a fine degree of accuracy and flatness on their measuring faces, that they may be "wrung" together and the overall length of any number of blocks so joined is the sum of their individual lengths. These gauges are often called Johannsen gauges after their originator. (Fig. 7) The Coventry Gauge and Tool Co. Ltd. Fig. 7 A set of slip gauges (107 pieces). {Lid of box not shown) The number of blocks in a set will depend upon the range of sizes that is required to be made up, and a medium-sized set of such gauges contains 47 pieces of the following sizes. 18 MEASUREMENT AND GAUGING Metric dimensions: widths of bloeks in mm. Pieces Range Steps 1 1005 — 9 101-109 0-01 9 1=1-1-9 01 24 1-24 1-0 4 25 , 50, 75, 100 25-0 47 Total From such a selection it is possible to choose blocks to build up almost any dimension that can be named within the capacity of the set. It will be seen that there will be alternative methods of making up a size, but the one should be chosen which employs the smallest number of pieces. Example 3. Choose blocks to assemble the following sizes: (a) 37.31 mm 0)6 1-685 mm. To assemble (a) we may use the following blocks: 101 1-3 1000 2500 37-31 Dimension (b) may be obtained as follows: 1005 108 1-6 8-00 50-00 61-685 The above example should indicate to the reader how to proceed for any other size, the method being to start at the figure on the extreme right of the dimension required, choosing gauges to accommodate each figure in turn. English sizes For routine English work, slip gauges made to inch dimensions may be obtained but it is possible, by calculating the metric equivalent, to use gauges from a metric set. Example 4. Make up a set of slip gauges to check a gauge measuring 2- 1758 in. THE SPIRIT LEVEL 19 The metric equivalent of 2-2758 in is found by multiplying 2-1758 by 25-4 and is equal to 55-26532 mm. The nearest size to this, that may be assembled from the set of gauges given above, is 55-265 which is 0-00032 mm or 0-000012in. too small. Gauges to make up to 55-265 are as follows: 1005 106 1-2 2 50 55-265 Exercises 2b From the list given on p. 18 make up sets of slip gauges to give the following sizes: 1,11-11 mm 2. 23-64 mm 3. 35-635 mm 4. 68-78 mm 5. 75-70 mm 6. 11 5-36 mm 7. 9.52mm 8. 44.45mm 9. 80.99mm 34-925 10. A dimension given as , 4 q- mm is to be checked. Make up two separate sets of blocks, one to measure each limit. 11. Make up sets of slip gauges to check the jaws of a limit gap gauge for a 30mm BSI f7 shaft. 12. Two holes, 0-875 inch diameter and 10625 inch diameter, are bored in the face of a casting at 1-5625 inch centres. Convert to mm, and make up a set of slip gauges to test between the insides of test plugs placed in the bores. The spirit level The spirit level consists essentially of a glass vial fixed into a frame. The inside top surface of the vial is not straight but is formed to a radius, con- vex upwards, as shown in Fig. 8. In some instruments the inside of the vial is ground barrel-shaped as shown, whilst in others the glass tube which Fig. 8 20 MEASUREMENT AND GAUGING forms the vial is bent to a radius. The vial contains spirit with sufficient air space to leave a bubble, and is cemented into the frame and accurately located so that when the base of the frame is level the bubble rests at the centre of the scale. The relations governing the movement of the bubble and the angle of tilt of the level will be followed from Fig. 9 and the following: fait««t Bj_J ""77777777777777777777777777^77777777777^1^ Fig. 9 Arc CAD represents the upper inner surface of the vial and O the centre of its radius. OB represents the base of the level. OB is horizontal and perpendicular to OA. If OB in now tilted to bring B to B,, point A on the vial will swing to A, but the bubble will remain vertically above O and will travel along the vial to A. A,A If is the angle of tilt, then 6 (radian) = R If BB, is the length of the arc through which one end of the level (length BB, L) swings, then 9 (radian) = — jr± BB, L AA, R and BB, = L. AA, R Actually, the height h that one end of the level is above the other is the dimension we require, but when dealing with angles as small as those con- THE SPIRIT LEVEL 21 cerned here, the difference between BBj and h is so small as to be negligible. Thus we can say that h = ^-(distance bubble moves) If the angle of tilt is required in degrees, then since Movement of bubble (radian) = 9 (degree) = R 57-3 (Movement of bubble) R Unfortunately, not many makers of levels mark them with particulars as to the radius (R) of the vial, but this may be determined experimentally by tilting one end of the level by a known amount and after noting the move- ment of the bubble, calculating the radius from the above expressions. Example 5. A spirit level is 300 mm long, and it is found that when one end is raised 002 mm above the other, the bubble moves 1-50 mm along the vial. Calculate the radius of the vial. , 17 u . , . h Movement of bubble We have that y = = 0-02 _ 1-5 300 ~ R From which R = ^p^' 5 = 22 500mm or 22- 5m Example 6. The base of level is 450 mm long and the radius of the vial is 30 m. Find (a) the height of one end above the other, and (b) the angle of tilt, corresponding to a bubble movement of 3 mm. h _ Movement of bubble T = R * , _ L (movement of bubble) R 450 x 3 .... 0- 045mm 30 x 1000 Angle of tilt in degree = 3 q ? ^ ^qq^ = 0-005 73° = 20-6 second 22 MEASUREMENT AND GAUGING The sine bar For accurate work in connection with angles the sine bar possesses advantages over the usual forms of protractor. Sine bars differ in form, but the considerations affecting their setting are the same in every case. Two common types of sine bar are shown in Fig. 10. The bar shown at (a) has two plugs which are let in and project about 12 mm from the front face. At (b) is shown a bar which is stepped at the ends and a roller is secured into each step, being pulled in by a screw so as to contact with each of the faces of the step. Both at (a) and (b) the follow- ing points are important if the sine bar is to be of any use: (i) The rollers or plugs should both be of the same diameter. (ii) Their centre distance must be absolutely correct. (The diagram is dimensioned as 200 mm centres, but sine bars are available in 100, 250 and 300 mm centres as well. (iii) The centre line AB of the plugs must be absolutely parallel with the edge of the bar used for measuring (generally the bottom). It is desir- able for the two edges of the bar to be parallel, with AB parallel with both. Fig. 10. When in use, the bar shown at (a) lends itself to clamping against an angle plate, whilst that at (b) can be rested on two piles of Johannson gauges to give it the correct inclination. Calculation for sine bar setting In Fig. 11, C is the centre distance of the plugs, h is the height of one plug above the other and a is the angle to which the bar must be set. CALCULATION FOR SINE BAR SETTING 23 Fig. 11 Then QR h and h = C sin a i.e. difference in height of plugs = (centre distance) (sine of angle). Example 7. Calculate the setting of a 200 mm sine bar to measure an angle of 36° 38'. We have tljat sin 36° 38' = 0-5967 h = 200 X 0-5967 = 119.34 Hence one plug mu§t be set 11 9-34 mm above the other. Example 8. Calculate the setting of a 250 mm sine bar to check the angle of g taper of 1 in 16, ^n tjie diameter. The taper is shown in pig. 12(a) and, if it is assumed to pe 16 units long, then in triangle ABC, 16 Fig. 12(a) AC = 16, BC = 4, and BAC = ta 4-4 = 3i =o - 03125 24 MEASUREMENT AND GAUGING From which | = 1° 47' 30" and 6 = 3° 35' Now sin = sin 3° 35' = 0-0625 and since the setting is for a 250 mm sine bar: h = 250 X 0-0625 = 15- 625 mm Fig. 12(b) The set-up shown at Fig. 12(b), A and B being two sets of block gauges assembled to give the setting calculated above. Precaution when checking plane surfaces with the sine bar The reader should observe that when using the sine bar to check the angle between two plane surfaces (e.g. the surfaces of an angle plate) the bar must be set accurately at right angles to the slope of the face being measured. The following example will illustrate how an error may be in- troduced if this is not done. Example 9. A surface was being checked by a 100 mm sine bar which, due to an error in setting, was placed 6 mm out of square with the slope of the surface. The angle obtained from the sine bar readings was 59° 30'. Find the true angle of the surface being measured. The conditions are shown diagrammatically, and exaggerated, in Fig. 13. The readings should have been taken on the line of greatest slope AB, but were actually taken on AC. CH is a perpendicular drawn from C on to AB, and from the conditions of the problem, CH = 6 mm. AB == AC = 100 mm, E, F and G are points where horizontal lines through A meet verticals through B, C and H. GAUGING LARGE RADII 25 Fig. 13 In triangle ACH: AH 2 = AC 2 - CH 2 100 2 - 6 2 = 9964 AH = \/9964 = 99-82 Now since CH is parallel to the slope, C and H are the same height and CE = HG In triangle CAE: AC = 100, and CAE = 59° 30' .'.CE = 100 sin 59° 30' = 100 x 0-8616 = 8616mm Then since CE = HG; in triangle HAG HA = 99- 82 mm, HG = 86- 16 mm and HAG is the true angle of the plate Sine HAG = QQ _. = 0-8631 from which, HAG, the true angle of the plate = 59° 40' i.e. an error of 10' Gauging large radii From some classes of work it is necessary to measure the radii of circles which are too large to be straddled by calipers or a micrometer. Where the complete circle is available its circumference may be measured by a tape and the diameter obtained by dividing by n. This method, however, is by no means perfect, and is not at all convenient when the radius to be measured is not part of a complete circle. An alternative method is to determine the radius by reference to the distance of its surface from the corner of a vee block resting on it. 26 MEASUREMENT AND GAUGING In Fig. 14 ABC represents the faces of a vee block resting on a circle, centre O and radius R. r^'-i Fig. 14 If 6 is the included angle of ABC, its half angle ABO will be ■=• as shown. The circle and the block contact at D, and in triangle BOD, since D is a right angle, angle BOD = 90 - ^ For different-sized circles placed in the vee, the variable length that we can measure is BE, so that the problem becomes one of finding a suit- able expression for R in terms of this length and the angle of the vee. Now BE = BO - EO and EO = OD = R BO But OD = secant ( 90 - l) .\BO=ODsec (90 - |) = R cosec y Npte: secant = cos since sec ( 90 - -!)- cosec -7T Hence BE = i?cosecy- -R cosecant = -r— sin iacosecy— lj From which R = BE cosec GAUGING LARGE RADII 27 For any given angle of vee, the quantity cosec ^ - 1 is constant and can be calculated and stamped on the gauge. All that is then necessary is to measure BE, and divide it by this number to give R. Example 10. A gauge of the type shown in Fig. 14 having an included angle of 120° is placed on a tube and the length BE measures 2 1-25 mm. Find the diameter of the tube. Here R = BE 21-25 21-25 cosec y — 1 cosec 60° - 1 1-1547 - 1 21-25 0-1547 = 137-36 mm and dia. of tube = 2 x 137-36 = 274-72 mm The main difficulty in the use of a gauge of this type is the accurate measurement of the distance BE. This may be overcome by constructing the gauge and incorporating a depth gauge or micrometer head on the centre line at B. An alternative construction is to make the gauge with a fiat portion as shown at Fig. 15, using slip gauges to check the distance between the work and the flat. It will be an interesting example to plan such a gauge as this. Example 1 1 . Plan out a vee gauge to measure round work, to cover a range of diameters varying from 250mm to 750mm. Fig. 15 28 MEASUREMENT AND GAUGING We will make the gauge with an included angle of 120°, and the flat portion FG filling in the corner of the vee such that it just contacts with a 250 mm circle placed in the vee, as shown in Fig. 15 . The length of the gauge must be sufficient to accomodate a 750mm circle. The length must exceed twice the distance from E to line AB, i.e. > 2AE sin 30° = > 2 x 375 x h say, 400mm. If flat FG just touches a 250 mm circle placed in the vee, then CB = CD sec 30° = 125 x 1-1547 = 144.34mm and from the vee corner B to FG the distance is 144-34 - 125 = 19-34 mm. It now remains to find the relation between the radius of any circle placed in the vee, and the distance from its circumference to line FG. From Fig. 14 we have that from B to the circumference of a circle radius R, = R cosec j — 1 a which becomes 0-1547/? when j = 60°. We have to subtract 19-34 mm from this, so that d = 0-1547/? - 19-34 mm and transposing to give R results in D d+ 19-34 R = 0-1547 mm The distance d can be measured with slip gauges, or a micrometer depth gauge incorporated in the construction as before. By marking the above formula on the gauge, the checking of sizes becomes a routine job. The measurement of large bores An interesting example of the measurement of a large diameter is avail- able to us in the gauging of large holes with a point gauge . This is shown in Fig. 16(a), where a hole of diameter D is being gauged by a point gauge of length L. In practice, L is a very small amount less than D, and when the gauge is held at one end, a small amount of rocking movement on either side of the centre line is possible at the other end. This is indicated byw. THE MEASUREMENT OF LARGE BORES 29 Fig. 16(a) The conditions are shown exaggerated at Fig . 1 6(b) . The full circle is the hole being gauged, and the dotted circle (centre A) is that which the end of the point gauge would describe if it made a full sweep . Actually, the end of the gauge only moves over the arc BHF.The amount by which this guage is smaller than D is shown by CH. Let this be S. Fig. 16(6) An approximate solution to the problem which will be accurate enough for most practical purposes is as follows: 30 MEASUREMENT AND GAUGING If BC is joined, angle B is a right angle, since it is the angle in a semi- circle, and in triangle ABC: AC 2 = CB 2 + AB 2 . AC = D; AB = L and CB is very nearly equal to w. Hence we may say: D 2 = L 2 + w 2 approximately. But D = L + 8 (L + S) 2 = L 2 + w 2 L 2 + 2LS + S 2 = L 2 + w 2 Now S will be a very small quantity, probably less than 0-02 mm so that S 2 will be so small that we may ignore it. Hence U + 2LS = L 2 + w 2 2LS = w 2 S = 2L This enables us to find the amount the gauge is smaller than the hole when we know how far the end may be rocked on either side of the centre line. Example 12. If a 375 mm point gauge rocked 6 mm at one end, calculate the diameter of the hole being gauged. w 2 The difference (S) = ^ and if the total movement is 6 mm, w — 3 mm 3 2 9 ' ~ 0-012mm * 2 X 375 750 Hence, Hole diameter = 375-012mm Exercises 2c 1. The radius of the vial of a spirit level is 25 m. When this is placed on the bed of a machine 2^m long, the bubble is 3mm from its central position. What is the end to end error in the machine? 2. A machine bed is 1-8 m long and is tested by a level 150mm long. One division on the level corresponds to an inclination of 0-06 mm per m. The level is transversed in steps of its length, from the LH to the RH end of the bed, with the following results: Position of Level 1 2 3 4 5 6 Reading (Division) + LH end high - RH end high + k + 1 + 1* + i (continued on p. 31) THE MEASUREMENT OF ANGLES AND LARGE RADII 31 Position of Level Reading (Division) + LH end high - RH end high -1 10 U 11 -1 12 Make a scale diagram showing the dip in the bed to an enlarged scale, and calculate the total error. 3. If a level is to be sensitive enough to indicate 1 minute of angle by 3 mm of move- ment of the bubble, what must be the radius of the vial? 4. Calculate the setting of a 250mm sine bar to check a taper of 1 in 6 on the diameter. 5. A 100 mm sine bar is used to check the inclination of a surface which is given on the drawing as 26°36' ± 4'. The height of one plug above the other is found to be 45-03 mm. What is the error in the angle of the surface? 6. A sine square has 4 plugs spaced at the corners of a 125 mm square. To mark out the template shown at Fig. 17, the template is secured to the sine square and line AB is first marked parallel to two of the plugs 1 . The square is then tilted and set to the correct angles for marking BC, CD and AD. Calculate the settings for marking these lines and also the lengths x and v. Fig. 17 7. In testing an angular surface with a 200 mm sine bar, the reading obtained is 122-30 mm for the height of one plug above the other. If the bar is 6 mm out of alignment with the line of maximum slope, calculate the true angle of the surface and state the error in the reading. 8. A point gauge 500 mm long, when tried in a bore, rocks a total amount of 12 mm at one end. When it is tried in a position at 90° to the first position, the movement is 24 mm. Calculate the mean diameter of the bore and the out of roundness. 9. A vee of 120° with a 12mm flat at its bottom is placed on a cylinder and the distance from the flat to the curved surface of the cylinder is 2-40 mm. Calculate the diameter of the cylinder. 32 MEASUREMENT AND GAUGING 10. In Fig. 18 A and B are two spherical seating pins 14 mm diameter. Calculate the height H so that the distance between the circle and the setting pin C shall be 0-50 mm. (A and B are true half spheres.) 50 Fig. 18 11. How much rock must be allowed on a point gauge 460 mm long if the bore is to be finished to 460-06 mm diameter? The location of points on angular surfaces When a component having an angular surface is shown in orthographic projection on a drawing, only the projected view is given, and unless a true auxiliary view is added to give particulars of points located on the surface, some means must be found to calculate their position. Two or three problems of this type will be illustrated in the following examples: Example 13. The plan and elevation of a block with two holes is shown in Fig. 19 (a). Find (a) the centre distance of the holes, (b) the angle between XX and AB, both dimensions as measured on the sloping surface. In dealing with problems of this type it is well to cultivate the sense of visualizing lengths in 3 perpendicular directions, and also of being able to make a rough pictorial sketch of the data. The sketch is generally very useful in helping to show up how the problem should be treated. In Fig. 19 (b), A and B are the hole centres, C is a point on the sloping surface where the centre lines intersect, and D is the point where a vertical through A meets a horizontal through C. THE LOCATION OF POINTS ON ANGULAR SURFACES 33 - V ? (a) Fig. 19 (b) Then in triangle ACD: A A CD = 26 mm, D = 90° and C = 35° AC CD 26 cos 35° 0-8192 = 31-74mm The centre distance AB is the hypotenuse of the right-angled triangle ABC. AB 2 = 26 2 + 31-74 2 = 1684 AB = n/1684 = 41-04 mm The angle made by AB with the side of the block is the angle ABE and AE • ABC 26 AB = Sm ABE = 4T04 = 0-6335 /\ From which ABE = 39°18 / Example 14. Fig. 20 (a) shows the plan and elevation of a block in which a hole has to be drilled, entering at the point A, and leaving at B. Calculate the angular settings of the block for drilling. 34 MEASUREMENT AND GAUGING Fig. 20(a) Fig. 20(b) The centre line of the hole is lettered XY in the elevation, and X,Y, in the plan, point B being underneath the block as drawn. A pictorial view of the job is shown at Fig. 20 (b), and in that diagram AC is a vertical and BC a horizontal line. For drilling the hole, the base of the block must be set at an angle of 9, and the face containing A must be set at a, both angles being with the vertical. In triangle CDB: A CD = 29 mm, DB = 19 mm and D = 90° CB 2 = 29 2 + 19 2 = 1202 CB = ^1202 = 3467 mm tan = AC 1 1-5 CB 34-67 18°21' -0-3317 DB 19 ft ,_„ tan a = — = 29 = °- 6552 a = 33° 14' Example 15. Calculate the angular settings for drilling a hole on the line AB shown in Fig. 21 . If A is joined to C, triangle ABC is formed and AC is parallel and equal to the line DE, shown on the end of the bar. THE LOCATION OF POINTS ON ANGULAR SURFACES 35 DE = OEsec 45° = OE x 1-414 = 26 x 1-414 = 36-76 mm and in triangle ABC: BC + A 52 , „,„, AC = tanA = 36^76 =M146 from which A = 54° 45'. Hence, if lines DE and EC are marked on the bar, and the end of the bar set at 54° 45' to the horizontal with DE parallel to the line of greatest slope, a hole started at B will break through at A. Fig. 21 Example 16. In the drawing of the component shown at Fig. 22 (x)the two sloping holes starting at A and B must meet at a point 10 mm from the base of the block. Find the starting heights a and b of the angular holes. A diagrammatic view of the base of the block is shown at Fig. 22 (y) in which the lines AB, BC and AC are assumed to be horizontal ones (i.e. the projections of lines joining the points A, B and C). 36 MEASUREMENT AND GAUGING Fig. 22 In triangle ABD: AB 2 = 13 2 + 16 2 - 2 x 13 x 16 cos 60° (cosine rule) = 169 + 256 - 208 = 217 AB = v^217 = 1473 mm also AB 13 sin 60° sin B . a 13 sin 60° ._,.. sin B = — r-j-ss — = 0-7644 14-73 B = 49°51' A Hence A = 180° - (60° + 49° 51') = 70°9' /\ /\ Since DBC = 90°; ABC = 90° - 49°51 / = 40°9' /\ also BAC = 90° - 70°9' - 19°51' In triangle ABC; AB = 14-73. B = 40°9' A = 19°51' THE LOCATION OF POINTS ON ANGULAR SURFACES 37 also C = 180° - (19°51' + 40°9') = 120° BC 14-73 and - — irtog1 , = -: — r^rr from which BC = 5-77 mm sin 19°51' sin 120° also in the same triangle: AC 14-73 sin 40°9' sin 120° whence AC = 10-97 mm The centre lines of the holes of which AC and BC are projections, slope upwards at 20° and 30° respectively. Hence, height of A above C = AC tan 20° = 1097 x 0-364 = 399 mm and height of B above C = BC tan 30 = 577 x 05774 = 3-33 mm. This completes the solution of the problem and gives us the following data: dimension a = 3-99 + 10 = 13-99 mm dimension b = 3-33 + 10 = 1 3-33 mm Exercises 2d 1. A straight-edge is placed on a surface sloping at 36° 45' and is set at an angle of 15° to the line of greatest slope. Calculate the inclination of the straight-edge to the horizontal. 2. In Fig. 23 a hole is to be drilled, starting in the centre of the sloping face, and breaking out at the corner B. Calculate the angle between the vertical plane containing centre line AB of the hole, and the face BCD of the block. ^*^*^ D_^t^>^ ,/ civ? a V / ( l~ 51 Jf Fig. 23 3. For Fig. 23, calculate the angle between the centre line of the hole and the base of the block, for a hole starting at A and breaking out in the centre of the base. 38 MEASUREMENT AND GAUGING 4. In Fig. 24 A and B are two points on the sloping surface shown, and line AB is parallel to the sloping edge. Find (i) angle line AB makes with the horizontal, (ii) dimen- sions a and b, and (iii) distance AB. Fig. 24 5. In Fig. 25, 6 holes are equally spaced, start on an 80 mm pitch circle and break through on a 130 mm circle. Calculate the angle between the centre lines of one hole and the next. 6. In Fig. 26 find the distance a and the angle a for a hole whose centre line AB shall be tangential to the 60mm circle. 7. Calculate the height from the base to the centres of holes A and B in Fig. 27. 8. In Fig. 28 a hole AB starts at A and leaves the bottom of the block at B. Find the length of this hole and the angle it makes with the base of the block. Another hole is to start at H, and be drilled in a plane parallel to the plane CDEF. The second hole must run into the first (i.e. their centre lines must intersect). Calculate the angle that the centre line of the second hole must make with the base. Measurement of tapers by means of balls and rollers Male taper with rollers (Fig. 29 (a)). If two similar rollers are placed in contact with the taper on opposite sides as shown, then for rollers of diameter d and centers c: h = c tan (1) MEASUREMENT OF TAPERS BY MEANS OF BALLS AND ROLLERS 39 Fig. 25 Fig. 26 4 Holes equally spaced 165 f P.C.Crs. 36" 105 Vertical height from base Fig. 27 Fig. 28 40 MEASUREMENT AND GAUGING and the difference between the dimensions taken over the top and bottom pairs of rollers will be 2h. When the taper is dimensioned as 1 in a certain length (say 1 in /) on the diameter, then - = 2/ and A = 27 (2) Fig. 29 {a). A taper given in mm/ unit length may be converted to 1 in / by divi- sion. In some cases it may be necessary to have a check on the diameter D of the taper, and if this diameter is situated at a distance H from the centre of the top pair of rollers: In triangle BEF: BF a BE = COS 2' and since BF = A dia of roller = - 2 yop = cos y, from which BE 2 cosy The radius EL of the taper = -.- - GK (EG is parallel to the taper centre line). T -, T D _„, a D .,, a EL = -_- - EG tan ■_- = -=- - H tan y MEASUREMENT OF TAPERS BY MEANS OF BALLS AND ROLLERS 41 D. C (centre distance of rollers) = 2 (BE + EL) V 9 r»r»c In practice, C, d and H would be known, so we require to transpose for (d D „, a\ _ + __// tan y \ T / (3) This gives that and C „ A a d D -=■ + H tan T = -=- 2 2 „ a 2 2 cosy D = C+ 2#tan£ — 2 or cos T (4) If the taper is dimensioned as 1 in /: Then z / a 1 tan 2 = 27 and D then becomes a COS^r = 2 vTTT 2 (Fig. 29(A)). H /fv/T+72 D - C + — - * (5) Fig. 29 (6). In practice, for the measurement of tapers in this way, it is helpful to have a fixture of some kind which will support the taper and provide supporting and gauging arrangements for the rollers. 42 MEASUREMENT AND GAUGING Example 17. A taper of 1 in 10 on the diameter is 80 mm long. Two pairs of rollers 12 mm dia. are used to check it, and the spacing of the rollers is as shown in Fig. 30. If the reading over the top rollers is 105 mm, cal- culate (a) the reading for the bottom rollers, (b) the diameter of the taper on a circle 14 mm from the top. Fig. 30. If the top rollers are situated at 58 mm from the face on which the taper and bottom rollers are resting, the roller centres will be 52 mm as shown . Centre distance of top rollers = 105 - 12 = 93 mm and applying the formula to find h we have h = ^[where c = 52 and / = 10] 52 i *n = ~?j = 2-60 mm Hence the reading over the bottom roller will be 93 - 5-2 + 12 = 99-8 mm For calculating the diameter D of AB we have from (5) above that H dV L 4- I 2 and C - 93, H = 8, d = 12, / = 10 CHECKING A TAPER HOLE BY MEANS OF BALLS 43 So that 8 12VT+100 D ~ 93 + To 10 = 93-8 - 12-012 = 81 -788 mm Checking a taper hole by means of balls For this it is necessary to use two balls of different diameters which will rest in the hole, touching its sides. The ball sizes should be chosen to give a reasonable centre distance (c), and this may be measured by employing a depth gauge from the top face of the hole to the top of the lower bah\ and a height or depth gauge to the top of the upper one. Fig. 31 Generally, the dimensions R, r, c and h in Fig. 31 would be known, and we require to derive formulae for finding a and D. E is the point where the ball contacts with the side of the hole and BC is parallel to the side of the taper. Then in triangle ABC: AB = c y C = 90°, and B so that AC AC AB R — r (since CE = r), R . a sin j — r .a = Sin fr c 2 (6) which enables us to find the angle a of the taper. 44 MEASUREMENT AND GAUGING If the taper is dimensioned as 1 in / on the diameter, then from Fig. 29 (b), . a i sin^- = and from above, 2 ^TTT 2 R - r i vTTT 2 vJTT 2 = c 2(R - rY Square both sides: 4{R - r) 2 ' from which / 2 = c* (7) 4(i? - r) 2 4 and / = ,/ c2 - J c2 ~( R - V - Vc2 ~ ( R ~ r ) 2 2{R - r) ■ To obtain an expression for the top diameter D of the hole we must consider triangles AEF and FGH and D == 2(AF + GH) = 2^AE sec j + FG tan y J [since E = G = 90°] = 26? sec y + Atan f) (8) For a taper of 1 in / on the diameter sec j = * + F and tan | = 1 [Fig. 29 (6)] and D becomes 2 p ?V ' + /2 + Al (9) Example 18. In a check on a taper hole, using the symbols and method given in Fig. 31, the following results were obtained: CHECKING A TAPER HOLE BY MEANS OF BALLS 45 R = 15 mm, r = 12-5 mm, c = 37-4 mm and h = 7-3 mm. Calculate the total angle of taper and the top diameter. Here we have from above, that sin y = 15 - 12-5 ~ 37-4 This gives y = 3° 50' and a = 7° 40'. = 00668 D = 2(rscc^ + /itan^J sec^= 10022 tan j = 00670 D = 2(15 x 10022 + 73 x 00670) = 3 1-04 mm Exercises 2e 1. Calculate the diameter D and the included angle of the taper for the case shown in Fig. 32. Fig. 32 Fig. 33 2. For the gauge shown in Fig. 33 calculate the centre height H between the two balls. (See Example 22, p. 52.) 46 MEASUREMENT AND GAUGING 3. Calculate, for the set-up shown in Fig. 34, the dimensions A and B. (The rollers are all 20 mm dia.) 4. For the example shown in Fig. 35, make up an expression connecting H with the centre distance D of the rollers. Find H when D = 77 mm. Total. Taper 1 in 20 '777777777777?77777777)(7777777777777777777 58 <p -A Fig. 34 Two 20 Rollers Fig. 35 5. A hole A of diameter D has two equal plugs, B, which just fit into it. Obtain an expression in terms of D for the diameter d, of two more plugs C, which will just fit into the spaces left (Fig. 36). Find d when D = 80 mm. Fig. 36 Fig. 37 6. In Fig. 37 the three radii blend together. Calculate the angle a. WIRE MEASUREMENT OF SCREW THREADS 47 Wire measurement of screw threads For the fundamental accurate measurement of a screw thread such as is necessary in turning taps and screw gauges, the method employing 3 wires is a very useful one. The wires are arranged as shown in Fig. 38 (a). We will work out a general case, and in Fig. 38 (b) is shown a wire, radius r, resting in a sharp pointed thread of angle a, pitch/? and effective (or mean) diameter D E . The wire size is not important, providing the three are the same diameter, touch at the flat portion of the thread, and are large enough to project above the thread for gauging. The best results (a) Fig. 38 (*) are obtained, however, if the wires touch the thread at the effective diameter, and for this reason the wires should be near to the following diameters: For ISO Metric and Unified d = 0-577 p For Whitworth d = 0-564 p In Fig. 38 (b), AD AB cosec y = r cosec ■=- H DE cot y = f cot 7 CD = \H = 4 cot h = AD - CD = r cosec j - j cot -x- 48 MEASUREMENT AND GAUGING and distance over wires (W) = D E + 2h + 2r = d e + 2 ( r cosec f ~ f cot y) + 2r = D E + 2r(\ +cosec^-|cot| = D E + d\l + cosec 2)~J cot f ( wnere d = dia. of wires) Having established this general formula, which may be applied to any thread, we will determine its special adaptation for the most common thread forms. (a) ISO metric and unified (Fig. 39) Here D E = D - 2(0-325/>) = D - 0-65p a = 60° cosec y = 2 : coty = 1-732 W{over wires) 2 COt 2 = Z) £ + d (1 + cosec -y J = D -0-65p + 43)- | (1-732) = D + 3d - 1-516/j < J 6l 6 '^fr-^Vv 1 ' -p Fig. 39 ISO metric and unified Fig. 40 Whitworth CHECKING THE THREAD ANGLE OF A SCREW 49 (b) Whitworth (Fig. 40). Here the depth of thread is 0-64/7, so that D E = D - 0-64/? Also since a = 55° cosec-y = 2-1657 cot ^ = 1-921 fF(over wires) = D E + d(l + cosec -y] — y cot -y = - 0-64/? + </(3-1657) - L 1921 = Z) + 3-1657rf - l-60/> Example 19. Determine the measurement over wires for the following cases: (a) Af30 x 3-5 ISO metric using wires 2 mm dia. (b) 1 in dia. X 10 t.p.i. Whitworth using wires 0-062 in dia. (a) M30 x 3.5 ISO metric using 2mm wires. W = D + 3d - 1-516/? D = 30, d= 2 and/) = 3-5 .-. W= 30+ 3 x 2 - 1-516 X 3-5 = 30-694 mm (b) 1 in dia. x 10 t.p.i. Whitworth using 0-062 wires W = D + 3-1657rf- 1-60/7 D = \,d = 0-062 and/) = 01 W = 1 + 3-1657 x 0-062 - 1-60 x 0-1 = 1-036 in. Checking the thread angle of a screw By taking measurements over two sets of wires of different diameters a check on the thread angle may be made. The underlying theory of this is similar to that given on page 43 for the measurement of taper holes by means of two balls. 50 MEASUREMENT AND GAUGING Miscellaneous problems in measurement Examples in gauging and measurement are so many and varied that it is impossible to establish set rules for application to every problem. If the fundamental rules of geometry and trigonometry are known thoroughly it is nearly always possible to apply some of them to the solution of the problem. The following miscellaneous examples will serve to indicate possible methods of dealing with problems of a similar type. Example 20. Two 25 mm circles centres D and B, touch a 100 mm circle, and their centres subtend an angle of 50° at its centre. Find the diameter of a circle which will touch the other three. The problem is shown at Fig. 41 and we require to obtain the diameter of circle C . Fig. 41 If C and B are joined and triangle ABC is considered, we have: AC = 50 + r; CB = 12-5 + r, AB = 62-5 and A = 25° (r = radofC) Applying the cosine rule for triangle ABC CB 2 = AC 2 + AB 2 - 2AC.AB cos A" Substituting the values from above gives us (12-5 + a-) 2 = (50 + r) 2 + (62-5) 2 - 2(50 + r)(62.5)(0-9063) MISCELLANEOUS PROBLEMS IN MEASUREMENT 51 Multiplying out the brackets gives: 156-3 + 25r + r 2 = 2500 + lOOr + r 2 + 3906 - 5664 - 11329r eliminate r 2 from both sides and collect terms 25r - lOOr + 1 13-29 r = 2500 + 3906 - 5664 - 156-3 38-29r = 585-7 585-7 r = 38-29 = 15-29 mm giving a circle of 2 x 15-29 = 30-58mmdia Example 21. From a piece of round material 40 mm radius, a piece 50 mm radius is cut as shown in Fig. 42. Find the distance x . Fig. 42 If AC and BC are joined as shown in the figure, then in triangle ABC: AB = 59, AC = 40 and BC = 50 also AC 2 = AB 2 + BC 2 - 2AB.BC cos B (cosine rule) D AB 2 + BC 2 -AC 2 5-9 2 + 5 2 -4 2 ,. . , cos B = „ . ^ = — — — (working in cm) 2 AB.BC 43-81 2x5-9x5 59 A B = 42°3' If CD is 1 r to AB, then = 0-7425 DB = CB cos B = 50 cos 42°3' = 50 x 0-7425 = 37- 125 mm x = AB - DB = 59 - 37-125 = 21 -875 mm 52 MEASUREMENT AND GAUGING Example 22. A profile gauge is as shown by ABCD in Fig. 43. Two plugs are placed in the gauge and dimension h is required as a check. Fig. 43 In the diagram, EF and HM are drawn parallel to AB and PG in per- pendicular to it. HN is a horizonatl line, and if we can find GN we shall be able to determine h. CK 26 x tan 32° = 16-25 mm In triangle JCG: JCG = ^ = 61°, JG = 12-5 and JC = JG cot 61° = 12-5 cot 61° = 6-93mm In triangle ALB: LB = ALtan24° = 26 tan 24° = 11.57mm In triangle JGF: JF = JG tan 24° = 12-5 tan 24° = 5-56 mm FC = JC - JF = 6-93 - 5-56 = 1.37mm BF = LK - (LB + FC + CK) = 51 - (11-57 + 1-37 + 16-25) = 21-81 mm GP = BF cos 24° = 21-81 cos 24° = 20-01 mm PM = rad of small plug so that GM = 20-01 - 8 = 12-01 mm GH = sum of plug radii = 12-5 + 8 = 20-5 mm /\ GM cos HGM = -p^Y Grl 12-01 /\ 1Zr W1 = 0-5858 and HGM = 54°8' 20-5 /\ /\ /\ Now HGN = PGN - HGM and since PGN = 90° +24° = 114° .HGN = 114 - 54° 8' = 59° 52' GN = HG cos HGN = 20-5 cos 59° 52' = 20-5 x 0-5020 = 10-29 mm Distance h = [JG + GN + rad of small plug] - 26 = 12-5 + 10-29 + 8-26 = 4-79 mm MISCELLANEOUS PROBLEMS IN MEASUREMENT 53 Example 23. For the turned part shown in Fig. 44, calculate the radius (/?) so that the diameter at the throat may be 20mm as shown. Fig. 44 The radius R blends into the 20° angular portion, and into the 20 mm radius, spherical end. We have to construct and solve an equation to give us R. DC is drawn perpendicular to AB, and DF is drawn to the point where R blends into the 20° angular portion so that FDE = 20° . GFE is parallel to AB. Then we have that GF + FE + CB = AB = 51 mm. GF = GH cot 20° = (AH - AG) cot 20° = (AH - = [AH - (CD - DE)] cot 20° = [19 - {(/? + 10) - R cos 20°}] cot 20° = [19 - R - 10 + 0-9397/?] cot 20° = [9 -0-0603/?] cot 20° = 24-73 -0-166/? FE = /? sin 20° = 0342/? CE) cot 20 c CB = VDB 2 - DC 2 = V(R + 20) 2 - (/? + 10) 2 Hence equating (GF + FE + CB) to AB = 51 24-73 - 0-166/? + 0-342/? + V(R + 20) 2 - (/? + 10) 2 = 51 and this reduces to V(R + 20) 2 - (/? + 10) 2 = 26-27 - 0-176/? Square both sides: (/? + 20) 2 - (/? + 10) 2 (26-27 - 0-176/?) 2 54 MEASUREMENT AND GAUGING i.e. by squaring out the brackets R 2 + 40R + 400 - R 2 - 20R - 100 = 690-1 - 9 25R + 0-03LR 2 Collecting and re-arranging on the i?H side: 0-03 IR 2 - 29-25.R + 390-1 =* This is a quadratic equation in R, and can be solved by the formula method. R = 29-25 ±V29-25 2 - 4 x 0-031 x 390-1 2 x 0-031 From which R = 930 or 13-55 The smaller root, R = 13-55 is obviously the one we require Hence R = 13-55 mm Exercises 2f 1. Calculate the diameter over wires for the following screw threads: (a) M20 x 2-5 ISO metric over 0-577/>mm wires (b) M36 x 2 ISO metric over 0-577pmm wires (c) fin X lOt.p.i. Whitworth over 0-564/? wires 2. A screw thread has the form and angle shown in Fig. 45. Calculate the reading W over 8 mm plugs placed in opposite threads. •* J. 7&4 \Lj2 n pitch Fig. 45 3. A wedge rests between two radiused jaws as shown in Fig. 46. When the top of the wedge is level, calculate the distance H. 4. Solve the previous problem, when the radius on the right-hand side jaw is 3 mm. MISCELLANEOUS PROBELMS IN MEASUREMENT 55 5. In Example 3, if the wedge has an included angle of 40°, and is tilted through an angle of 10°, calculate the distance H to its higher corner. 6. Calculate the distance from the centre of the circle (O) to the centre of the 20 mm plug placed in the slot shown in Fig. 47. Fig. 48 7. Fig. 48 shows the profile of a die form. From the information given, calculate the width d at the narrowest portion. Fig. 49 8. From the information given in Fig. 49 calculate the width W of the profile shown. 3 Calculations for cutting, turning and boring Speed and feed range The reader will, no doubt, be acquainted with the meaning of speed and feed in connection with turning and boring operations. The driving arrangements of machine tools usually make provision for a number of speeds and feeds, so that a suitable one may be chosen for the work in hand. The reader will probably be curious as to how these are determined. In the case of spindle speeds, the highest and lowest speeds in the range are generally related to the extremes of size for which the machine is designed. For example: a lathe might be designed to take a rr.nge of work varying from 10 mm to 250 mm diameter. Allowing for a cutting speed of 22 m/min, this would give for the top speed: XT 1000 x 22 1000 x 22 x 7 _„ . . N = T7 — = ^ 77j = 700 rev/min n x 10 22 x 10 as being suitable for the 10mm diameter work. For the lowest speed: XT 1000 x 22 1000 x 22 x 7 „ . . N = » X 250 = 22 x 250 = 28 rev/min These will be the highest and lowest in the range, and if we assume there are eight speeds altogether, six intermediate speeds must be chosen and fitted in, the intermediate speeds being so calculated that the whole series is in some regular order. One method of arranging the speeds would be to make them in straight line form, in which each speed would be the same amount greater than the one below it. In this case, as there are 8 speeds and 7 intervals, each interval would be: Top speed — bottom speed _ 700 — 28 _ 672 _ _, 7 ~ 7 - — - *& SPEED AND FEED RANGE 57 and the speeds would be: 1st 28 rev/mm, 2nd 28 + 96 = 124 rev/min 3rd 124 + 96 = 220 rev/min and so on. 700 - 3 4 5 6 Speed Number Fig. 50 If these were plotted on a graph, the result would be as shown in Fig. 50, and the series is known as an Arithmetic Progression. In practice, speeds arranged in this way are not suitable, as the steps between the speeds at the lower end (28 rev/min, 124 rev/min, 220 rev/min) are too great, whilst at the upper end of the range (700, 604, 508 rev/min, etc.) a larger interval value could be tolerated without inconvenience. To overcome these objections and provide a convenient range of speeds, they are generally arranged in Geometric Progression. When arranged in this way, instead of each speed being a constant amount greater than the one below it, the speed is a constant multiple of the one below it. The calculation for determining speeds arranged in geometric progression is as follows: Considering the case we have taken, where the extremes are 28 and 700 rev/min with 8 speeds. The 2nd speed will be a constant amount multiplied by the 1 st, and the 3rd will be the same constant multiplied by the 2nd, and so on. Let this constant be denoted by K. Then 1st speed 2nd speed 3rd speed 4th speed = 28 = 28 X K = 28AT = 28*: x K = 2SK 2 = 2SK 2 x K = 2SK 3 and so on to the 8th speed, which we can see will be 2%K\ 58 CALCULATIONS FOR CUTTING, TURNING AND BORING Now the 8th speed is 700 rev/min, so that K 2SK 1 = 700 28 ZD No -- v/25 = 1-584 25 Log. 7)1-397 9(0-1997) Hence we have 1 st speed = 28 rev/min 2nd speed = 28 x 1-584 = 44 rev/min 3rd speed = 44 x 1-584 = 70 rev/min, and so on. The speeds arranged in geometric pregression are shown plotted in Fig. 5 1 , and their comparison with the straight-line arrangement is shown in the table below. The reader will observe that the geometric arrangement gives closer intervals at the bottom of the range and wider ones at the top which is more useful under application. 700- Arithmetic Geometric Speed Progression Progression rev/min rev/min 1 28 28 2 124 44 3 220 70 4 316 111 5 412 176 6 508 279 7 604 442 8 700 700 3 4 5 6 Speed Number Fig. 51 We might put a geometric progression in general symbolic form as follows: Let A x = 1st term A 2 = 2nd term A n = nth term K = constant multiplier n = number of terms SPEED AND FEED RANGE 59 1 st term = A , 2nd term = A 2 = A X K 3rd term = A 3 = A 2 K = A,K 2 nth term = A n = A X K" - l (i.e. if n = 8, then the index of K is 8 - 1 = 7). Hence if A n =A,K n ~ x *»- 1 = d* and £ = J4 5 - In an example, A, , A n and n would be given and the expression above would enable K to be found and the whole series calculated. Example 1 . Calculate a suitable range of six speeds for a drilling machine, if the size range of the machine is to be from 2-5 mm to 10 mm drills and a cutting speed of 22 m/min is to be given. Show a speed table with suitable drill size for each speed. IQO0S We have that N = — -3— where jV = Spindle speed (rev/min). S = Cutting speed (m/min). d = Drill diameter. For the top speed (2-5 mm drill) „ 1000 X 22 1000 x 22 x 7 0Qnn . . N * = *X2.5 = 22 x 2-5 = 2800rev/min Lowest speed (10 mm drill) „ 1000 x 22 1000 x 22 x 7 „_ . . Nl = .xlO = 22 x 10 = 70 ° rev/mm fa 72800 AT (the multiplier) =Jf =a/^ = >/4 No. Log. 4 5)0.6021(0-1204 Antilog 0-1204 = 1-319 = K. We may now calculate the range of speeds as follows: 1st speed = 700 rev/min 2nd speed = 700 x 1-319 = 924 rev/min 3rd speed = 924 x 1-319 = 1220 rev/min 60 CALCULATIONS FOR CUTTING, TURNING AND BORING 4th speed = 1220 x 1-319 = 1610rev/min 5th speed = 1610 x 1-319 = 2120rev/min 6th speed = 2120 x 1-319 = 2800rev/min The relationship between the speeds and the drill sizes they will accom- modate may now be calculated: 1st speed, 700rev/min suitable for 10 mm drill 700 2nd speed, 924rev/min suitable for 10 x q~j 700 3rd speed, 1220rev/min suitable for 10 x 4th speed, 1610rev/min suitable for 10 X 5th speed, 2120rev/min suitable for 10 x 1220 700 1610 700 2120 Nearest 0-25 mm 10mm 7-5 mm 5-75 mm 4-25 mm 6th speed, 2800rev/min suitable for 2-5 mm drill The speed and drill diameter table is shown below. 3-25 mm 2- 5 mm Spindle Speed (rev/min) 700 924 1220 1610 2120 2800 Suitable drill diameter (mm) 100 7-5 5-75 4-25 3-25 2-5 Feeds The factors governing the choice of feed-range limits is rather beyond our scope, but when the limits of the range have been fixed, together with the number of intermediates in the range, the steps usually follow the rules for geometric progression in the same way as the speeds. Exercises 3a 1. A lathe is operating on a range of work varying from 25 mm to 250mm diameter. Allowing for a cutting speed of 22m/min, calculate the highest and lowest speeds necessary. If there are 8 speeds in the complete range, find the range of speeds if they are in geometric progression. Make out a table showing the most suitable diameter to be turned on each speed. 2. A drill has 4 speeds and drills a range of holes from 2 mm to 6 mm diameter. Calculate the four speeds if they are in geometric progression, and make a drill diameter speed table. Cutting speed = 16-5m/min. CUTTING TOOL LIFE 61 3. For Ques. 1 p. 60 plot a graph showing spindle speeds vertical, and speed number horizontal. 4. On a 400mm stroke single-pulley, all-geared shaping machine the number of turns of the driving pulley required to make one complete double stroke of the ram were found to be as follows: 1st speed, 27 turns 2nd speed, 16 turns 3rd speed, 8 turns 4th speed, 4 turns a • *u *• Cutting time 1-25 Assuming the ratio -=— — ^—. to be — ; — and to remain constant, estimate a Return time 1 ' suitable pulley speed to give an average cutting speed of llm/min in the lowest gear and on the longest stroke. With this pulley speed, find the most suitable stroke for each of the other speeds. 5. The highest spindle speed for a small lathe is 1500 rev/min. In order to obtain a suitable cutting speed for drilling some 3 mm holes in brass, a drill head is mounted on the carriage, and driven in the opposite direction to the spindle. At what speed must the drilling spindle be driven to give a cutting speed of 66 m/min? Cutting tool life As a cutting tool does its work it becomes blunted, and a time ultimately arrives when it must be taken out and re-sharpened. The life of the tool between the times of re-grinding is influenced by the severity of its treatment whilst it is cutting. Depending upon circumstances, there is a best economic tool life for every tool; if the cutting duty is made such as to allow the tool to last longer than the best economic time, then it is cutting below an efficient rate and is doing less work then it might. On the other hand, if its performance is raised to a level such as to cause it to become blunted in less than the economic time, then undue ex- pense and lost time are being incurred in the additional sharpening and re-setting necessary. The problem of tool life and of cutting generally is rather complicated and indeterminate, since there are so many variable factors involved. From experimental data, however, cutting speed and tool life have been found to conform roughly to the following rule: VT» = C, where V = Cutting speed in metres per minute T ~ Corresponding life in minutes C = A constant depending on cutting conditions 62 CALCULATIONS FOR CUTTING, TURNING AND BORING \ to \ for roughing cuts in steel \ for roughing cuts in cast iron > i for light cuts in steel ) — 12 _ J_ — 10 for roughing cuts in steel j ~ :j ^ ;„„,„ carbide tools. The above values of n are only approximate and are influenced by tool shape, use of cutting compound, etc. The above relationship enables us to estimate probable tool life as shown in the following examples: Example 2. When operating with roughing cuts on mild steel at 20m/min a certain tool gave a life of 3 hours between re-grinds. Estimate the life of this tool on similar cuts at a speed of 30m/min. In the VT" = C expression for this case we will take n = £. The first step is to calculate the value of the constant C. We have that when V = 20, T = 180 min and n = \. Hence C = 20 x 1801 Taking logs No. log T „ , _ log 180 180 8)2-2553 LogC=log20+ 8~ 180r ^02819 = 1-5829 20 1-3010 1-5829 Antilog 1-5829 = 38-27 = C. Hence we may write: We now require T when VT* = 38-27 V= 30 307^ - 38-27 T k = 2*£Z = 1.2757 T = (1-2757) 8 T = antilog 0-8456 = 7-008, say 7 min. No. Log 1-2757 0-1057 8_ 0-8456 TOOL CUTTING ANGLES 63 Example 3. A tool cutting at 20m/min gave a life of 1 hour between re- grinds when operating on roughing cuts with mild steel. What will be its probable life when engaged on light finishing cuts? [Taken = £ for rough- ing, and fo for finishing cuts.] Here we have for roughing: No. log 20 x 60i = C 60 8)l-7782( C = antilog 1-5233 = 33-38 60i 0-2223 20 1-3010 1-5233 Applying to the finishing conditions 20 x n = 33-38 n - 3 ^ 8 - 1-669 T = (1-669) 10 No. log 1-669 0-2225 (1-669) 10 2-225 T = antilog 2-225 - 167-9, say 168 min. Tool cutting angles The principal angles on a cutting tool are its rakes and clearances, and these are shown in Fig. 52. Side Rake Front (Top) Rake Front Clearance Side Clearance Fig. 52 The choice of suitable cutting angles depends upon the material being cut, and the reader should look up particulars in books dealing with 64 CALCULATIONS FOR CUTTING, TURNING AND BORING Workshop Technology. We might consider here the effect on the cutting angles of various tool settings. When a tool is cutting a circular bar of material it is operating relative to a radial line drawn from the centre of the work to the cutting point. Thus in Fig. 53(a) the tool is cutting relative to the line OA. If the tool point is is level with the centre of the work the line OA is horizontal and the cutting angles operating are the true values put on the tool. If, however, the tool point is placed above or below the centre, the cutting angles will be modified since the line OA is not now horizontal. Top Rake (r) i_ Clearance(c) Fig. 53(a). Top Rake ^■Clearance Fig. 53(6). In Fig. 53 (b) the tool point is shown a distance h above the work centre. The line OA is now tilted up an angle a, and if R is the radius of the work we have that sin a = -=- K The top rake angle of cutting will be increased by the angle a, and since the total angle /5 has not changed the clearance will be reduced by a. If the tool is put very high, the clearance will be so reduced that it will vanish altogether and the tool will rub instead of cut. If the tool is placed below the centre the effects are opposite, the rake being decreased and the clearance increased. We might calculate the effect on the cutting angles of a numerical example. Example 4. A bar of material 60 mm diameter is being turned with a tool having 20° top rake and 6° front clearance. Calculate (a) the cutting angles when the tool is 2-5 mm above centre, and (b) the amount the tool must be above centre for the clearance to become zero. TOOL CUTTING ANGLES 65 The conditions are as shown at Fig. 53 (b). (a) sin ^^^J^ 0-0833. From which a = 4° 47'. The rake is increased and the clearance decreased by this amount. Thus the rake becomes 20° + 4° 47' = 24° 47' and the clearance be- comes 6° - 4° 47' = 1° 13'. (b) If the clearance is to vanish, then angle a must be 6°. sin a = ^7j-and since sin 6° = 0-1045 0-1045 = *j- h = 30 x 01045 = 3-14mm Exercise 3b 1. A certain tool when cutting cast iron had a life between regrinds of 2 hours when cutting at 20m/min. If the relationship between life and speed is given by VT™ = C, cal- culate C, and estimate the tool life at a speed of 15m/min. 2. For the tool in Question 1, plot a graph of tool life-cutting speed over a range of speeds from 30m/min to 15m/min. 3. For a certain tool it was found that the relationship between speed and tool life was given by VFi = 50. Estimate the cutting speed to give a time of 2 hours between re- grinding. 4. If the relationship for high-speed steel tools is VT* = C„ and for tungsten carbide tools VT 5 = Q, and assuming that at a speed of 25m/min the tool life was 3 hours in each case, compare their cutting lives at 35m/min. 5. For a certain tool it was found that the relationship between cutting speed (K)and tool life (T) was as follows: Express T in terms of V and find T when V = 25m/min. 6. A cutting tool has a top rake of 20° and a clearance of 7° . Calculate the modified values of these angles when the tool is cutting 3 mm below centre on a bar 44 mm diameter. 7. When turning a bar 50 mm diameter, how much above centre may a tool with a clearance of 6° be set before the clearance vanishes? When the tool is in this position, what is the effective value of the top rake, if the rake on the tool is 15°? 8. A boring tool 10mm deep is required to bore out holes to 40mm diameter. If the body of the tool is horizontal, how much clearance will be necessary if the bottom corner of the tool is to clear? 66 CALCULATIONS FOR CUTTING, TURNING AND BORING 9. If the clearance face of the tool in Question 8 is made with two angles as shown in Fig. 54, calculate the angles a and /5 for the tool to clear the hole. Fig. 54 10. A 50 mm diameter boring bar is concentric with a 72 mm hole which is being bored. The tool is 12mm square and passes through the centre of the bar. Calculate the tool angles necessary so that the cutting rake shall be 10°, and the clearance 6°. Taper turning When a tapered or formed surface is being turned the accuracy of the sur- face is influenced by the position of the tool point relative to the centre of the work. The conditions in the case of taper work are shown in Fig. 55 (a) . If the tool is set on the centre, and its movement controlled so as to turn the correct taper, it starts at A, and when it has travelled the line AB, the correct taper has been produced. End View Fig. 55 If, now, instead of being on the centre, the tool is a distance h below centre and starts at C, it will travel along the line CD, and if D is on the same circle as B, the length CD is greater than AB. The tool being set to TAPER TURNING 67 move the distance AB will therefore not reach the point D, and con- sequently if A and C are on the same circle, the top diameter of the taper will be less than it should be. The reader will probably appreciate the point better if he considers the extreme case of the tool being at E and travelling parallel to AB. In such a case, if the tool could cut in this posi- tion, it would not turn a true taper, but would produce a tapered shape faintly hollow in form. In order to follow the problem mathematically we will show an enlarged diagram (Fig. 55 (6)). The tool starts at C and moves out the distance CF, equal to AB. If it turns the end of the bar to radius r, instead of reaching the radius OB = R it will only attain OF = R l . Let the work be L long as shown in Fig. 55 (a). True taper = -* — p — * Actual taper obtained = *■ ' ~ — - An attempt to reduce R , to an expression in terms of the other quantities involves rather awkward terms, and cases will be best evaluated from the information available . Such an example is illustrated as follows: Example 5. In turning a taper of 1 in 6 on the diameter, the tool is set to move the correct angle relative to the work, but is 4mm below centre. If the work is 32 mm diameter at the small end, calculate the actual taper obtained. Referring to Fig. 55 (b), if we assume the work to be 60 mm long, then the large diameter will be 32 + 10 = 42 and we shall have r = 16 mm R = 21mm h = 4mm The tool, starting at C where OC = 16 mm, will move out to F where CF = 21mm. We require the distance OF. OF 2 = FG 2 + OG 2 = (FC + CG) 2 + OG 2 But FC = 5 and OG = 4 .-. OF 2 = (5 + CG) 2 + 4 2 68 CALCULATIONS FOR CUTTING, TURNING AND BORING But CG 2 = CO 2 - OG 2 = r 2 - h 2 = 16 2 - 4 2 240 CG = v / 240~ = 15-49 Hence OF 2 = (5 + 1549) 2 + 4 2 = 2049 2 + 4 2 = 435-8 OF = \/43T8~= 20-88 The top diameter of the taper will therefore be 2 x 20-88 = 41-76 instead of 42mm. Since the bottom diameter is 32 mm and length 60 mm Actual taper = 41-76 - 32 = 9^6 H 60 60 This gives a taper of 1 in 6-15. Form tools For turning forms from the cross-slides of turret and automatic lathes and sometimes from centre lathes as well, a tool is used which gives the correct form on the work. If the tool is set on centre as shown in Fig. 56, it must have the correct form in plane OAB. Since, however, the tool must have clearance as shown, lengths such as BC, taken perpendicular to the front clearance face will be less than lengths such as AB taken on the horizontal, and the form of the tool on a plane parallel to BC will be different from its form on AB . When the tool is being made, the shaping and other machining operations are carried out parallel to the clearance face, so that for the purpose of making the tool it may be necessary to determine its form when taken on a plane such as BC, perpendicular to the front clearance face. U ^Clearance Fig. 56 FORM TOOLS 69 We shall best illustrate the method of determining the modified form of the tool in planes perpendicular to the front clearance face by working one or two examples: Example 6. A tool is to have the form shown in Fig. 57 (a) on its top horizontal face. If the clearance is 10°, determine and sketch the form on a plane perpendicular to the front clearance face. 15 10 72 ^ i. 55 - (a) 11 2 15 10 7-; i ^ p / V r 55 » e) ( IF2> Fig. 57 In Fig. 57 (a) horizontal dimensions will not be affected, but vertical BC dimensions will be shortened in the ratio of -j-tt in Fig. 56, and since /\ gp ABC is equal to the clearance angle, — — = cos of clearance angle AB = cos 10° = 0-9848 Hence the 11-2 in dimensions become 11-2 x 0-9848 = 11-03 mm and the 45° angle will become a triangle as shown in Fig. 57 (b). tan A = -^- = 1-0154. From which A = 45° 26' 11-03 Hence the revised sketch of the tool profile taken along the clearance face will be as shown in Fig. 57 (c). When the form of the tool incorporates circular shapes the problem becomes rather more involved, since the effect of shortening the depth but not the width converts the circular form into a portion of an ellipse. 70 CALCULATIONS FOR CUTTING, TURNING AND BORING Example 1 . Calculate and sketch the form of the tool shown in Fig. 58 (a) when taken perpendicular to the front clearance of 12j° . As before, the vertical dimensions are shortened in the ratio of the clearance angle cosine, i.e. cos 12|° = 0-9763. We thus have 12-8 x 0-9763 = 12-50 mm 8 x 0-9763 = 7-81 mm and for the 15 mm radius, 15 X 0-9763 = 14.64mm The base of the 20° angle is shortened to 0-9763 of its original length, so that if we divide the tangent of 20° by 0-9763 we shall have the tangent of the modified angle tan 20° = 0-364 Q976 3 = 0-3728 = tan of modified angle = tan 20° 27' The sketch of the modified profile is shown at Fig. 58 (b) . 12-5 116 30 88 6 1 2-8 8 ^C20 c iM 75 (a) 12-5 11-6 30 _8_-8 6 Elliptical Form Fig. 58 The finishing of the elliptical form given to the circular portion is apt to be troublesome, but such a shape can be produced on a grinding wheel by trimming it with a radius forming attachment set off centre. This is shown in Fig. 58 (c), and if the radius truing attachment is set with the FORM TOOLS WITH "TOP RAKE" 71 diamond rotating in plane CDE it will trim the wheel to a semicircle, 15 mm rad., in that plane. Since AB is less than CD, the true form of the wheel on a radial plane such as ABO will be elliptical, because the width of the wheel at B = width at D. The semi-minor axis of the elliptical form will be AB = FD, and the major axis will be 2 x 15 mm = 30 mm. As the profile on a radial plane is the one imparted by the wheel to the work we can, by forming the wheel in this way, obtain the required elliptical shape for the tool in question and we require to determine h in order that when CD = 15 mm, FD will be 14-64 mm. Let us assume a grinding whell of 200mm radius, and consider the problem from the aspect of two intersecting chords of a circle. Then FD.DH = CD.DG But FD = 1464, DH = 400 - 14-64 = 38536 and CD = 15 Hence (14-64) (385-36) = 15.DG „ 14- 64 x 385-36 „, , mm DG = rr = 376-1 mm and CG = 3761 + 15 = 3911 391-1 CE = ±CG = ^- = 195-55 But h 2 = R 2 - CE 2 = 200 2 - 195-55 2 = 1760 from which h = 42 mm Hence by trimming the wheel to a 15 mm radius in a plane 42 mm off- centre, the required elliptical form will be produced. Form tools with "top rake" If back slope is put on the cutting face of a form tool the cutting effect of the tool is rather curious, because for the purpose of obtaining an accurate relative reproduction of the tool form on the work, the tool, at the finish of the cut, must have its top face lying on a radial line. This is shown at Fig. 59 (a), where the tool is shown at the completion of its cut, and its top face lies on the radial line OA. The reader will notice that at this position the effective top rake is zero, since the tool is cutting relative to tangent BC, which is perpendicular to OA. When this tool starts its cut, however, the conditions are as shown at (b), and it will be seen that if the back slope a on the tool is made large 72 CALCULATIONS FOR CUTTING, TURNING AND BORING enough, the tool will start cutting with an effective top rake of/3 = a - S. As the tool feeds in, this rake will gradually get less until, as we have seen above, there is zero rake at the final position. The reader will notice that a tool of this type must be set below the centre by the amount h, where h - = sin a and r = smallest radius being turned. The calculation for the modified form to which the front face of such a tool should be made is similar to that we have already dealt with, except that the angle a must be taken into account. In Fig. 59 (c), a is the back-slope angle and c the clearance. The tool finishes its cut relative to face AB, but the depths of the form are put on parallel to AD (i.e. along CB). Hence a length AB on the top of the tool will correspond to CB, per- pendicular to the clearance face. In triangle ABC: C = 90° and A^Sb = a + c CB = AB cos (a + c). Hence depths in the form must be shortened in the ratio of cos (a + c). CIRCULAR FORM TOOLS 73 Fig. 60 Circular Form Tool. Circular form tools On some types of automatic lathes forming is done from the cross-slide by means of a circular form tool, a sketch of which is shown at Fig. 60. These tools have the advantage that the form may be turned on their rim and they may be used all round the rim by continual re-sharpenings. Cutting clearance is obtained by making the cutting edge AB some distance h below the centre, and the tool is applied to the work as shown at Fig. 61 . The clearance angle a is then the angle CAO and sin a = AO Tool radius (r) and h = r sin a Fig. 61 74 CALCULATIONS FOR CUTTING, TURNING AND BORING Gashing the tool in this way results in a variation between the form turned on it and the form it imparts to the work, because a radial depth DB on the tool will turn a depth AB on the work. Widths on the form are unaffected, and corrections for depths may be calculated as follows, where d = a depth on the tool and / = corresponding depth on work. From the property of intersecting chords of a circle But and Hence DB.BF = ABBE. DB = d; BF = 2r - d; AB - / BE = AE — / = 2r cos a — I d(2r - d) = l{2r cos a - I) which reduces to a quadratic equation in d as follows: d 2 - 2rd + l(2r cos a - I) = in which all the quantities except d are known. The following example will illustrate the application of this. Example 8. A circular form tool, 100 mm diameter, is to be made to produce the form shown in Fig. 62 (a). If the gashing is to give a cutting clearance of 10°, determine the form to be turned on its periphery. 7-5r^ m 425- (b) Fig. 62 Here r --= 50 mm and a = 10°, so that h = 50 sin 10° = 50 x 0-1736 = 8-68 mm. To turn the correct diameters on the work, the steps on the face AB of the tool must be: 32-5 - 17-5 __ ,32-5 - 12-5 ,. ■= = 7-5 mm, and ■= = 10 mm Applying the equation above to these two cases, we have CIRCULAR FORM TOOLS 75 (0 / = 7-5, r = 50, cos a = cos 10° = 0-9848 d 2 - lOOtf - 7.5(100.0-9848 - 7-5) - from which d = 7-38 mm («) / = 10, r = 50, and cos a = 0-9848 d 1 - lOOrf - 10(100.0-9848 - 10) = from which d = 9-82 mm The angle on the tool to give an included angle of 1 20° on the work must now be corrected . Its width is 32-5 - 12-5 tan 30° - 10 tan 30° = 5-774 mm and its corrected depth from (ii) above is 9-82. Hence 5-774 noy = tan of its angle — 0-5882 from which the angle is found to be 30° 28'. The turned profile of the tool is shown at Fig. 62 (b). Exercises 3c 1. A lathe centre is being ground up by a 70 mm diameter grinding wheel fixed to a tool-post grinder on the compound slide. If the centre of the wheel is set 10mm below the axis of the lathe centre, and the compound slide fed at 30°, find the angle to which the centre will be ground. 2. A lathe is set correctly for turning a taper of! in 6 when the tool is on the centre. What taper will be produced on apiece ofwork 120 mm long and 60 mm top diameter when the tool is 5 mm below (jentre? 3. A form tool is straight, and set at an angle of 15° with the axis of the work (i.e. to form an included angle of 30° on the work) . If the tool is set to the above angle, but 5 mm be- low centre, calculate the actual angle produced on a job 50 mm top diameter and 40 mm long . ?7Z7 -&. I Fig. 63 4. A form tool having 8° clearance and 15° back slope is required to turn the diameters shown at Fig. 63. Calculate (a) the depth AB on the clearance face of the tool, (b) the amount the tool should be below centre, (c) the top cutting rake at the commencement of cutting. (Top diameter ofwork = 10mm). 76 CALCULATIONS FOR CUTTING, TURNING AND BORING 5. Calculate the tailstock set-over for turning a taper of 8° included, on a job 115 mm long. What variation in taper will be produced by a variation of ±2-5 mm in the length of the work? 6. Calculate the depths and the angle on a form tool for producing the form shown in Fig. 64. Tool has no top rake and 10° clearance. Fig. 64 7. At Fig. 65 is shown a thread form which is to be finished by the form tool indicated. Calculate the angle, depth and bottom land as measured from the clearance face of a tool having no top rake and 15° clearance. 5-24 Fig. 65 8. A circular form tool 55mm diameter is gashed to give 10° clearance, and is to turn the form indicated at Fig. 63. Calculate the amount off-centre for the gashing and the depth AB on the tool. 9. Calculate the depths and angle to be turned on a circular form tool 60mm diameter to give the profile shown at Fig. 66, if the tool is gashed to give 12° clearance. APPROXIMATE CHANGE WHEELS FOR ODD THREADS 77 375^ \8-76 Fig. 66 10. In cutting the thread on a core for a die-casting mould, the pitch of the thread is to be made 2% longer than standard to allow for contraction. To do this, the tailstock of the lathe is set over, the lathe taper attachment is set parallel to the work, and then the thread is cut in the usual way. Calculate: (a) The actual pitch required if the nominal thread is 2mm pitch. (b) The tailstock set-over if the core is 150 mm long between centres. (c) The angle to which the taper attachment must be set. Calculating approximate change wheels for odd threads Sometimes a case may arise where an odd thread must be cut, the exact pitch of which cannot be obtained with the standard machine change- wheels. Also, if a lathe is not supplied with the special 127T wheel, and a metric pitch is required, some alternative way of getting a suitable pitch becomes necessary. The method of continued fractions will often provide a very near ratio to that required, and enable a pitch to be cut which is near enough for the purpose. In Appendix VII it will be seen, that the convergents of a continued fraction are a series of fractions, each succeeding one approaching closer to the true value of the original ratio. If, therefore, we have a complicated ratio, the exact value of which cannot be obtained on the machine, it is quite possible that by converting to a continued fraction and finding the convergents, one of these convergents will be a ratio that can be used, and its value will probably be close enough for the purpose. The method will be best illustrated by examples. 78 CALCULATIONS FOR CUTTING, TURNING AND BORING Example 9. Find the nearest pitch to 2- 18 mm that may be cut on a lathe with a 5 mm leadscrew, and give suitable change wheels from a set ranging from 20T to 120T in steps of 5T. ^, . r Drivers .„ , 2-18 mm 109 , . irkn . The ratio of gears: -p-^ will be — E = ^~r, and since 109 is a 6 Driven 5 mm 250 prime number, the exact ratio could not be obtained without a gear of this size. Converting this to a continued fraction and finding the convergents we have: 109)250(2 The continued fraction is: 218 I 32)109(3 2+1 96 3 + 1 13)32(2 2+1 26_ 2 + 1 6)13(2 11 1)6(6 and the convergents are: 1st = \, 2nd = f, 3rd = &, 4th = & 5th = $§. If we take the 4th convergent (#) we may obtain a gear ratio as follows: n _ 2 x 8-5 _ 20 x 85 Drivers 37 ~ 6 x 6-5 ~ 60 x 65 Driven To find the actual pitch obtained we must multiply the ratio $ by the pitch of the leadscrew, i.e. $ X 5 = ff, which when converted to a decimal gives 2- 1795 mm. This is less than 0-03% in error on the required pitch of 2- 18 mm. Example 10. Find the nearest pitch obtainable to 2\ mm, on a lathe with a 6 t.p.i. leadscrew and a set of wheels as in the last example. 2Jmm The pitch of the leadscrew is £in and the ratio required Converting the inches to millimeters (lin = 25-4 mm) 2\ 6 9x6 27 270 = 24 x nn '25-4 * 254 4x254 50-8 508 POWER REQUIRED FOR CUTTING 79 The continued fraction is 270)508(1 270 1 + 1 238)270(1 1 + 1 238 7+1 32)238(7 2+1 224 3 + 14)32(2 28 4)14(3 11 2)4(2 and the convergents: 1st = }; 2nd = ±; 3rd = &; 4th = #; 5th = ft; 6th — 2io Ulll — 508 The 4th convergent is the last one which can be made into a ratio and gives: 17 = 2 x 8-5 20 x 85 Drivers 32 ~ 4x8 ~ 40 x 80 Driven The actual pitch obtained will be £ x £ X 25-4 = 2-2489 mm, being 0001 1 mm short. Power required for cutting Turning and Boring. When metal is being cut with a single-point tool as in turning and boring, the tool is subjected to pressure in three directions at right angles: (1) vertical chip pressure, (2) horizontal work pressure across the lathe, (3) horizontal feeding pressure along the lathe. The first of these is of greatest importance from the aspect of the power absorbed. The other two, although absorbing some power, are of small effect when compared with the vertical pressure and are generally neglected. From numerous experiments that have been made it has been esta- blished that the cutting force on a single point tool is connected in an expression of the form F = Cd a f b where F = force; d = depth of cut;/ = feed, and C = a constant. a and b depend on the metal being cut and other factors. 80 CALCULATIONS FOR CUTTING, TURNING AND BORING For most practical purposes the expression F = Kdf= K(Cut area) gives results good enough. K is a constant depending on the metal being cut. If S ~ cutting speed in metres per minute, the work done per minute F x S will be F x S, and the power — ^ — Watts. Hence Power = 60 qq Q Kilowatts Approximate values for K are as follows: Metal Steel Steel Steel Steel r „ t A , being 100-150 150-200 200-300 300-400 f^T Brass Bronze 2\„ cut Brinell Brinell Brinell Brinell ™*~ K (N/mm 2 ) 1200 1600 2400 3000 900 1250 1750 700 [From the form of the expression, the reader will observe that K is the force on the tool per square millimetre of cut area.] When the power required to do the cutting has been calculated, the total power to run the cut and overcome friction in the machine may be found by adding on about 30%. Example 1 1 . Calculate the power being absorbed in running a cut 3 mm d^ep with a feed of 1-5 mm, on a mild steel bar 50 mm diameter turning at 140rev/min. . n X 50 x 140 Cutting speed = - 1000 22 x 50 x 140 ~ 7 x 1000 If we take the constant K as 1200 = 22m/min D 1200 x 3 x 1-5 X 22 Power = 60000 = 198kW Adding 30% for frictional losses in the machine we have 198 + 1-98 Xi|= 1-98 + 0-594 = 2-574 say, 25 kW to run the machine . POWER FOR DRILLING 81 Power for drilling When a drill is cutting it has to overcome th e resistance offered by the metal and a twisting effort is necessary to turn it. This effort is called the Turning Moment or Torque on the drill . The units for torque are those of a force multiplied by a length and the most usual is the Nm unit. The turning effect of a force, or a pair of forces, acting at a certain radius, is found by multiplying the force by the radius, or for two forces the turning effect is the sum of the product of each force by its radius. Thus in Fig. 67, if the drill required a torque of lONm to turn it, the torque would be equivalent to equal and opposite forces of 250 N each operating at 20 mm radius. [T = 250 x 0-02 + 250 X 0-02 = lONml SON Resistance at each corner Fig. 67 In addition to the torque, a drill requires an axial force to feed it through the work, but in power calculations this is generally neglected. When the torque is known, the work done is found by multiplying it by the number of turns made and by In . Thus if T = torque in Nm and N = speed in rev/min, the work done per minute = In NT Nm , , 2nNT , „ r and the power = ^^ kw 82 CALCULATIONS FOR CUTTING, TURNING AND BORING The torque required to operate a drill depends upon various factors, but for the purpose of being able to obtain an approximate calculation for it we will omit all but the drill diameter, the feed and the material being drilled. The relation between the torque, the diameter and the feed has been found experimentally to be that torque varies as/ " 75 /) 1 ' 8 . Using this, we may say that Torque (T) = Cf°- 15 D hS newton metres where C = a constant depending on the material / = drill feed (mm/ rev) D = diameter of drill (mm). When the torque has been found, the power can be calculated as shown above. The following table gives approximate values for the constant (C). A , . . . , . , .,, , Alu- Soft Cast Steel Carbon Material being drilled , , ., , s , minium brass iron (mild) tool steel C 0-11 0-084 007 0-36 0-4 Example 12. Calculate the power required to drill a 20 mm hole in mild steel at 250rev/min and a feed or 0-5 mm/ rev. Find also the volume of metal removed per unit of energy. Taking the constant, C, from the table as 0-36 we have 7 , = 0.36/°- 75 Z) 1 - 8 Nm / =0-5 D = 20 T = 0-36 (0-5)°- 75 (20) 18 and taking logarithms. log T = jog 0-36 + 0-75_log 0-5 + 1-8 log 20 = j-5563 + 0.75(1-6990) + 1.8(1-3010) = 1-5563 + 0-75(-0-3010) + 1-8(1-3010) = -0-4437 - 0-2258 + 2-3418 = 1-6723 T = antilog 1-6723 = 47 Nm Since the speed is 250rev/min _ 27r.250.47 _ 60 000 ~ 1ZJKW POWER FOR TURNING AND DRILLING 83 Volume of metal removed per minute = (Area of hole) (Feed) (Speed) = j x 20 2 x 0-5 x 250 = 39 275 mm 3 Energy consumption = — mow = 31-9mm 3 /watt minute. = 0-53mm 3 /joule Exercises 3d 1. Calculate the nearest change wheels for cutting a sparking-plug thread (l|-mm pitch) on a lathe with a 4 t.p.i. leadscrew and a set of wheels ranging from 20T to 120T in steps of 5T. For the ratio you select, find the actual pitch of thread obtained. 2. A worm having a lead of Timm is to be cut on a lathe with a 5 mm leadscrew. Taking n as 3-1416, express the ratio required as a continued fraction, and find the nearest convergent that can be used with a set of wheels specified for the last example. What was the actual error in the lead obtained for the worm? 3. A shaft revolves at 15 rev/min and requires a thread cutting on it which will cause a nut to move along the shaft at 66-5mm/min, when it turns at the above speed. Find the lead of the thread required and calculate the nearest that can be cut to it on a 5 mm lead- screw with change wheels specified for Question 1. What is the actual speed of the nut with the thread you obtain? 4. Estimate (a) the power input to a lathe when it is taking a 6 mm cut in cast iron with 0-75 mm in feed at 20m/min. (b) The volume of metal removed per unit of energy. Take the overall efficiency of the machine as 70%. 5. A lathe is just able to run a cut of 5 mm depth at a feed of 0-8 mm in steel of 120 Brinell, at 150 rev/min on a 50 mm diameter bar. Estimate what cut could be taken on 25 mm bars of 250 Brinell material, at 240 rev/min and the same feed as before. [Take values of K from the table on p. 80.] 6. Taking the value of K from the table on page 80, estimate what cut could betaken on a lathe turning bronze b ars at 0-6 mm feed and 20 m/min, if 5 kW were available, and 30% of the power were lost in friction. 7. For the lathe in Question 4, estimate the power cost per 8-hour day, with power at 2p per kWh and the efficiency of the motor is 80%. 8. Calculate the torque required to drill 20 mm diameter holes in mild steel, at a feed of 0-25 mm/ rev. If the drill speed is 300 rev/min, calculate (a) the power absorbed in cutting, (b) the energy absorbed per cubic millimetre removed per minute. 9. A 25 mm drill is drilling aluminium at HOm/min. Calculate its speed. If the feed is 0-3 mm/rev, calculate the torque, and the input power if frictional losses are equivalent to 30% of the cutting power. 10. For the drill in Question 8, calculate the drilling time for 100 holes, each 40 mm deep. If the electrical efficiency is 80%, calculate the cost to drill these 100 holes using the same electrical data as in Question 7. 11. By using a continued fraction calculate the nearest set of change wheels to cut the thread specified in Example 10(a) Exercises 3 c, leadscrew 5 mm pitch. 4 Calculations for gears and gear cutting Formatiion of the involute tooth For various practical and theoretical reasons, the tooth shape most commonly used for gearing is the involute. Before we commence our con- sideration of various problems connected with involute teeth, it will be as well to examine the involute curve itself. Involute If a cord is wrapped tightly round a circular form and then unwound, at the same time being kept tight, the end of the cord will trace out an involute. This is shown at the top of Fig. 68, where CB is the portion of the cord that has been unwound and AC is the involute. Another method Involute Involute Straight- edge Fig. 68 of tracing an involute is to roll a line (e.g. a straight-edge) on a circle when the end of the line will trace out an involute. The size of circle will, of course, influence the shape of the involute, but there are certain properties which are common to all involutes. Those properties which are interesting from the aspect qf the involute as a tooth form, are as follows: INVOLUTE TOOTH FORM 85 (1) A tangent to the involute is always perpendicular to a tangent from the same point on the involute to the circle from which it is formed. In Fig. 68 the cord CB is tangential to the circle at B, and a tangent to the involute at C is perpendicular to CB. (2) The length of the cord CB is equal to the length of the arc AB. Tooth form O, and 2 are the centres of a pinion and gear of which the pitch circles are shown tangential at P, which is called the pitch point . AB is a line through P perpendicular to line C^Cv CD also passes through P and is included at the angle if> to AB. CD is called the line of action and $ the pressure angle. The name "line of action" is given to CD because it is on, and along that line, that the pressure between the teeth takes place. The angle (j> nowadays is virtually always 20°, but at one time 14£° was more common, being related to half the inclined angle of the Acme thread form . With the changeover of the pressure angle from 14£° to 20° the clearance between the top of the tooth of one gear and the base of its mating tooth form has been increased from 5% of the circular pitch ( = 0-157 of the module), to 0-25 of the module. The cutting depth for gears of 20° pressure angle is thus 2-25 times the module. To Fig. 69 Formation of Involute Tooth. 86 CALCULATIONS FOR GEARS AND GEAR CUTTING obtain the tooth shape a circle is drawn tangential to CD called the base circle. The portion of the tooth between this circle and the top (EF) takes the form of an involute to this circle; the portion of the tooth below this circle (FG) is radial (i.e. on a line joining the end of the involute to the gear centre). This is all shown in Fig. 69, where for the sake of clearness the construction has only been carried out on the lower gear. The con- struction also only shows one side of a tooth; the other side is merely the same shape reversed, and spaced away, a distance equal to the tooth thickness. The involute rack The rack is a gear of infinite diameter so that its pitch circle will be a straight line (AB, Fig. 70). A base circle of infinite diameter tangential ^ F Fig. 70 to the line of action CD will be a straight line coinciding with CD. The involute to this will be the straight line EP, and the radial continuation will be PF. Hence the side of the rack tooth is straight, and inclined at the pressure angle. The complete tooth will have a total angle equal to twice the pressure angle (Fig. 70). In the following considerations of gear elements the following symbols will be used for the quantities stated: No.ofteethingear ... Tor r Pressure angle (ji Diametral pitch P Module m Circular pitch p Addendum of tooth . . Add. Diameter of pitch circle Dord Dendendum Ded. Radius of pitch circle . R or r The recommended manner of quoting the size of a gear tooth form, irrespective of the units employed, is to quote the addendum and refer to it as a module. In which case, pitch circle diameter = module x number of teeth or D = mT THE TOOTH VERNIER 87 The module is therefore the reciprocal of the diametral pitch, irrespective of whether measurements are made in millimetres or inches. All the formulae which follow can be used for diametral pitch, by substituting — for m. Until the use of the metric module becomes the preferred usage in describing gear tooth sizes, the reader may find gears listed in a diamet- ral pitch series. The diametral pitch is simply the number of teeth per unit of pitch circle diameter, and consequently it is necessary to be par- ticularly careful in stating that unit, e.g. diametral pitch of 8 or diametral pitch of 0-2 (inch series) (mm series). In any case, conversion to a module is simply effected from the relation- ship that the module is equal to the addendum; furthermore circular pitch p = n x module m The tooth vernier The gear-tooth vernier is an instrument for measuring the pitch line thickness of a tooth. It has two scales and must be set for the width (w) of the tooth, and the depth (h) from the top of the tooth, at which w occurs (Fig. 71). Fig. 71 The angle subtended by a half tooth at the centre of the gear (AOB) in Fig. 71). . ,360 90 = iof^=^=- 88 CALCULATION FOR GEARS AND GEAR CUTTING m> w *ri ■ 90 z> • 90 AB = y = AO sin ~y = R sin -=■ D = module x number of teeth D = 2R = mT a n mT and R = -j- „ w _ . 90 mT . 90 Hence -^ = i? sin -=- = -z- sm -= A T ■ 90 and w = wisin-=r To find fc we have that h = CB = OC - OB. But OC = R + Add. =^ + m A ™ Z> 90 W7, 90 and OB = R cos -=- = -=- cos -~. TT r mT mT 90 .... ,, — ■ ^ Hence /* = -y- + /w =- cos -^ = m + -^-| 1 - cos -^ \. (2) mrf 90l T-[ ! - CQ8 TJ Example 1 . Calculate the gear tooth vernier settings to measure a gear of 33 T, 2-5 metric module ^ . 90 ._ ,, . 90 w = mT sm-jr = 2-5 X 33 sin^- = 82-5 sin 2°43' = 82-5 X 0-0474 = 3-91 mm , mT.. 90, h = m + -2~(1 - cos yr) = 2-5 + 2 ' 5 * 33 (1 - cos2°43') = 2-5 + 41.25(0-0011) = 2-545 mm Constant chord method One drawback to the method just outlined is that the measurements w and h depend on the number of teeth (T) in the gear, and for each dif- ferent gear a fresh calculation has to be made. The following method avoids this and gives a constant pair of readings for all gears of the same pitch and pressure angle. CONSTANT CHORD METHOD 89 Fig. 72 In Fig. 72 is shown a gear tooth meshing symmetrically with a rack. O is the pitch point, and as we have seen above, the gear tooth will contact with the straight-sided rack tooth at the points B and D, lying on the line of action. Then DB = w will thus be constant for all teeth of the same pitch and pressure angle. Since EOA is the pitch line of the rack nm and EA = j circular pitch = \p = -y- OA = !EA = ™ A A In triangle OAB: B = 90° and O = <f> :. OB = OA cos i/j, and in triangle OCB: C = 90° and B = ^ .\CB= OB cos ij). nm Hence CB = OA cos iji cos ijt = OA cos 2 i/j = —r- cos 2 iff and 7tm (3) DB = 2CB = w = -j- cos 2 (/) h = Add. - OC = m - OC. OG = OB sin ip and OB - OA cos iff. OC = OA cos (/t sin $ = -j- cos ^ sin ^ nm . . . f~ 7t , • ,1 — j— cos (j> sin iff = m\ 1 — -j cos $ sin ty j It will be seen that expressions (3) and (4) remain constant if module (m) and pressure angle (^) do not alter. But Hence and h = / (4) 90 CALCULATION FOR GEARS AND GEAR CUTTING Example 2. Calculate the constant chord and the depth at which it occurs, for a 30T gear of 6mm module, 20° pressure angle. Here we have h> = —x- cos 2 20 sin 20° = 0-342 cos 20° = 0-9397 6 X 3-142 X (0-9397) 2 = 8-32 mm m(l — -j cos ij) sin f) = 6(1 - 0-7854 x 0-9397 x 0-342) = 6 x 0-7476 = 4-49 mm Plug method of checking for pitch diameter and divide of teeth The tooth vernier gives us a check on the size of the individual tooth, but does not give a measure of either the pitch diameter or the accuracy of the division of the teeth. Fig. 73 Fig. 73 shows a rack tooth symmetrically in mesh with a gear tooth space, the curved sides of the gear teeth touching the straight rack tooth at the points A and B on the lines of action. O is the pitch point. If now we consider the rack tooth as an empty space bounded by its outline, a circle with centre at O and radius OB would fit in the rack tooth and touch it at A and B (since OA and OB are perpendicular to the side of the rack tooth). Since the rack touches the gear at these points, the above circle (shown dotted) will rest against the gear teeth at points A and B and will have its centre on the pitch circle. In triangle OBD: OB = radius of plug required. CHECKING FOR PITCH DIAMETER AND DIVIDE OF TEETH 91 OD = \ circular pitch = A A B = 90° O = ijf. Tim 7im OB = OD cos if/ = -j- cos (jf nm Dia of plug = 20B = -y- cos iji (5) This is the diameter of a plug which will rest in the tooth space and have its centre on the pitch circle. Notice that the plug size remains the same for all gears having the same pitch and pressure angle. With such plugs placed in diametrically opposite tooth spaces, it is a simple matter to verify the gear pitch diameter. The accuracy of the spacing over any number of teeth may be found as shown in chordal calculations. Example 3. Calculate for a 36T gear of 5 mm module and 20° pressure angle, (a) plug size (b) distance over two plugs placed in opposite spaces, (c) distance over two plugs spaced 10 teeth apart. Tim ^S7T (a) Dia of plug = -^ cos ^ = ~- cos 20° = 7-854 x 0-9397 = 7-38 mm Pitch dia of gear = mT = 5 x 36 - 180mm (b) Distance across plugs in opposite spaces = 180 + 7-38 = 187-38 mm (c) Distance across plugs spaced 10 teeth apart (Fig. 74). Fig. 74 92 CALCULATION FOR GEARS AND GEAR CUTTING 'lf l {\ Angle subtended by 10 teeth = 10 x -~y- = 100° . In triangle OAB: AB = OA sin 50° = 90 x 0-766 = 68-94 Centre distance of plugs = 2 x AB = 2 x 68-94 = 137- 88 mm Distance over plugs = 137-88 + 7-38 = 145-26 mm Base pitch The base pitch is the circular pitch of the teeth measured on the base circle. It is useful for checking the angle between adjacent teeth and for checking a tooth against "drunkenness." Fig. 75 If AB (Fig. 75) represents a portion of the base circle of a gear, and CD and EF the sides of two teeth, and then the length FD is the base pitch. But if any lines such as CE and HG are drawn tangential to the base circle cutting the involutes at the points shown, then EC = GH = FD If there are T teeth in the gear, then T x FD = 2ttR b and FD iTlRt If ij) is the pressure angle, then from Fig. 69, page 85, H0 2 = rad. of base circle (R B ) = P0 2 cos ij> = R cos tjt. BASE PITCH 93 Hence FD = base pitch = ,£° S ^ n , 2nR TtD But ~y~ = ~f = 7tm ( D \ (since -= = m ) Hence base pitch = nm cos <ji (6) This is the distance between the curved portions of any two adjacent teeth and can be measured either with a height gauge or on an enlarged projected image of the teeth. Exercises 4a 1. Determine the diameter of a plug which will rest in the tooth space of a 4mm module 20° rack, and touch the teeth at the pitch line. Calculate (a) the distance over two such plugs spaced 5 teeth apart, (b) The depth from the top of the plug to the top of the teeth. 2. Calculate the gear tooth caliper settings for measuring the following gears: (a) 37T, 6mm module; (b) 40T 20mm circular pitch. 3. A 5 mm module involute rack tooth is measured at its pitch line and found to be 7-99 mm wide. If the tooth spacing and angle are correct, what error has been made in the cutting of the teeth? (Pressure angle = 20°.) 4. A 30T replacement gear of 5 mm module is required, and the nearest cutter available for cutting the teeth is one of 5 diametral pitch, (inch series). If the blank is turned to the correct module dimensions, and the cutter sunk in to the depth marked on it, what will be the error in the tooth? 5. Determine the "constant chord" dimensions for the following gears: (a) 8 mm module, 20° pressure angle; (b) 25 mm circular pitch, 20° pressure angle. 6. Calculate the diameter of plug which will lie in the tooth space of a 5 mm module gear with its centre on the pitch circle. If the gear has 50T, find (a) distance over two such plugs spaced in opposite spaces, (b) distance over two plugs spaced 12 spaces apart (u = 20°). 7. Two plugs are placed in adjoining spaces of a 29T gear 20mm circular pitch, and the gear is stood up resting on them. Calculate the distance from the face upon which these plugs are resting, to the top of a similar plug placed in the tooth space at the top of the gear. [Press, angle = 20°, and plug diameter is that which rests with its centre on the pitch circle.] 8. Determine the base pitch of the following gears: (a) 30T 8 mm module, ij> = 20°; (b) 30T, 25 mm circular pitch, ij> *= 20°. 9. Two teeth of a 30T gear of 1-25 mm module, 20° pressure angle are projected to a magnification of 50. Calculate, to the nearest 0-5 mm the length on the projected image of the following measurements: (a) base pitch, (b) depth of tooth space, (c) chordal thickness of tooth at pitch line, (d) height from root of tooth to pitch line. 94 CALCULATION FOR GEARS AND GEAR CUTTING Stub teeth For some purposes, particularly when gears are subject to shock and vibration, the tooth of standard proportions is apt to be weak and liable to break. In such cases stub teeth are often used. For this type of tooth the size is indicated by a fraction. The numerator of the fraction expresses the module to which the circumferential pro- portions of the tooth conform, and the denominator determines the radial proportions of the tooth. Thus a £ stub tooth means one in which the pitch diameter is worked out on a basis of 5 mm module, and the tooth height on the proportions of 4 mm module. Since 5 mm module gives a larger normal tooth than 4 mm module, the result is a short stubby tooth, hence the name. Stub teeth are generally cut with a pressure angle of 20°, and the following are the pitches most commonly used: 5 6 8 10 12-5 , 15 and 4' 5' 6' 8 ' 10 12-5 The same method is used when quoting stub tooth sizes in diametral pitches, but in this case the ratio produces a "proper vulgar fraction", i.e. with the numerator smaller than the denominator. Hence, if a stub tooth is denoted by a fraction in which the numerator is smaller than the denominator, the reader should appreciate that the sizes quoted refer to diametral pitches and not modules, and it will be necessary to state whether the diametral pitch is "millimetre series" or "inch series". Example 4. Calculate the principal dimensions for a 45T gear having a f stub tooth. Here: the pitch diameter will be worked out on 6 mm module and the heights on 5mm module. Pitch dia = mT = 6 X 45 = 270mm Addendum = m = 5 mm Top dia of gear = 270 + 2(5) = 280 mm Cutting depth = 2-25 X 5 = ll-25mm Backlash in gearing If a pair of gears were cut theoretically correct and assembled at the correct centre distance, a tooth on one gear would just fit hard into the tooth space of the other, because the pitch line width of the tooth and MEASUREMENT OF BACKLASH 95 space would be equal. For freedom of action the above conditions would be unsuitable, and it is usual to allow a little play between the thickness of the tooth and the width of the space into which it fits. This play is called "backlash," and it is the backlash which allows one gear to be turned a fraction before the drive is taken up by the mating gear. The amount of backlash to be allowed depends on the tooth size, and the following table gives an indication of suitable allowances: Module (mm) 10 8 6 5 4 3 2-5 2 1-5 Backlash [in mm clearance between face of mating teeth] 0-4 04 0-4 0-3 0-2 015 015 01 0-1 Measurement of backlash Two suitable methods of measuring backlash are (a) by means of feeler gauges between the teeth, (b) by measuring the distance that the centres of the gears may be moved nearer together from the standard distance before the teeth are in hard contact. The first of the above methods is straightforward and needs no mathe- matical manipulation. For the second method it will be helpful for us to obtain an expression giving the backlash in terms of the amount the gears are capable of being moved together. Fig. 76 In Fig. 76, ab is half the backlash and P is the pitch point. If we consider the portion cb of the tooth as being a straight line, then triangle abc is rightangled at b, ac is the amount the gear centers may be moved together, ab = \ backlash and angle acb = t/i. 96 CALCULATION FOR GEARS AND GEAR CUTTING Let B = backlash = lab and D = amount the gear centres can be moved together = ac- Then — = sin ib ac B_ 2 D = sin ij) and B = 2D sin (b B or D = 2 sin (Ji Helical (spiral) gears The type of gear we have dealt with so far has been the spur gear, i.e. one in which the teeth are straight and parallel with the axis of the gear. In helical gears the teeth are not straight, but are cut on a helix. These gears are also called spiral gears, screw gears and skew gears. A sketch of a portion of a helical gear is shown in Fig. 77, in which it will be noticed that the teeth slope at an angle a (the helix (spiral) angle) to the axis of the gear, and in the gear shown, the teeth are cut RH helix. Teeth may, of course, be RH or LH. Referring again to the figure, it will be seen that the pitch of the teeth may either be taken round the rim of the gear, or it may be taken perpendi- cular to the teeth. In the first case it is called the circumferential pitch ip c ), and in the second case it is known as normal pitch (p„). Circular pitch is still considered as being taken along the surface formed by the pitch circle. The circular pitches p n and p c are shown in Fig. 77, and as will be seen, the relation between them is the same as the relation between the A A sides AC and AB of triangle ABC. In this triangle C = 90° and A = helix angle a . Hence £*■ = — — = cos a, Pc AB i.e. p n = Pc cos a (7) Pn or p c = r = p n sec a rc cosct rn It will be seen that if a section is taken through the teeth on a plane con- taining AC, we shall have the true shape of the tooth as it is cut. Hence the normal pitch (p n ), which is the one measured parallel to AC, is the pitch which governs the cutter to be used to cut the gear . Referring again to Fig . HELICAL (SPIRAL) GEARS 97 [Measured orn [Pitch Circle] \ Fig. 77. 77, the pitch p c , multiplied by the number of teeth, gives the pitch circum- ference. This pitch, then, governs the size of the gear, which does not depend on the number of teeth. The larger the helix angle <r, the greater will be the ratio — and the larger the gear for a given number of teeth. Pn As well as circular pitch p„ and p c we may, just as in the case of spur gears, express the pitch in the module form . The normal module (m n ) is the one which governs the true shape of the tooth, and since cutters are most commonly specified in terms of the module, this is important to us from the point of view of cutting the gear. The relation between /w n and/>„ is the same as for spur gears. i.e. Pn = *™n (8) For a spur gear we have D = mT For a helical gear, since the circular pitch Pc = Pn COS<7 andD = , then m c = mT cos a cos a m cos a (9) 98 CALCULATION FOR GEARS AND GEAR CUTTING Addendum of the tooth = m and the cutting depth (working depth + clearance) == 2 addendum + clearance = 2 m + clearance. (10) In modern practice, using a 20° pressure angle, the cutting depth has been standardised at (working depth + 0-25 addendum) ==2°25/m (11) Helix (spiral) lead and angle The teeth of a helical gear are cut on a helix, which is merely a screw thread with a very large lead. The reader will, no doubt, be aware that a screw thread can be developed into a triangle, and the only differences between the development of a gear-tooth helix and that of a screw thread are: (a) the helix angle of a thread is the complement of that for a wheel tooth (Fig. 78(a) [complement of an angle = 90° - the angle] ; (b) the development is ' Helix ^^" Fig. 78 Development of Helix. based on the assumption that the helix makes one complete turn round the cylinder upon which it is cut. In a screw thread this is true, but the tooth of a gear only completes a small proportion of a complete turn. [The student may imagine a helical gear to be a short length of a very coarse CUTTER FOR HELICAL GEARS 99 thread having as many "starts" as there are teeth in the gear.] On the assumption noted under (b) above, the development of a helical gear tooth is shown in Fig. 78(6). From the diagram we have the following relationships Pitch circum nD tan a = = -j = -p- Lead L , , _ nD nmT sec a .,,-. or lead L = = — (12) tan a tan a Cutter for helical gears When cutting spur gears with a form cutter on the milling machine, the cutter is marked with the range of teeth for which it is suitable. For example, to cut a 30T gear, we should use a No . 4 cutter, which is suitable for a range of 26T to 34T. Due to the twist on the teeth of a helical gear this must be modified, and the size of cutter is given by No. of teeth (T) ( . (cos a) 3 ( ' Example 5. The pitch diameter of a helical wheel is to be approximately 120 mm, the helix angle is 30° and it is to be cut with a cutter of 4 mm module. Find the particulars of the nearest gear to this. We have that the normal module = 4 mm and a = 30°. 4T Hence, from (9) 120 = and T = cos 30° 120 cos 30° 4 = 30 x 0-866 = 25-98 The nearest to this is 267 , n mT 26x4 104 1oni and D = , ^r- = = 120-1 mm cost? cos 30° 0-866 Top dia. = 120-1 + 2 Add = 120-1 + 2(4) = 128-1 mm Cutting depth = 2-25m = 9-00 mm v mom a fur 7lD 3-142x120-1 ,.. From (12) Lead of helix = ^r- = ttt^ = 654 mm. tan 30° 0-577 100 CALCULATION FOR GEARS AND GEAR CUTTING Thus the nearest gear has the following particulars: 26T. Pitch dia. 1201 mm, Helix angle 30°, Lead of helix, 654mm. 96 The cutter for this gear would be ,_ , „ , = 40, i.e. the same cutter as (U-oOOJ would be used for a 40T 4mm module, spur gear. Example 6. Two parallel shafts at 120 mm centres are to be connected by a pair of helical gears to give a speed ratio of 1:2. The helix angle of the wheels is to be approximately 20° . If the normal module is 2-5 mm, deter- mine suitable wheels. Since the shafts are parallel, the helix angles of the two wheels will be the sarnie, but one will be RH helix, and the other LH Centre distance =120 Ratio of speeds = ratio of pitch radii =1:2 r = i x 120 = 40 and R = f X 120 = 80 Hence D = 160 and d = 80, also m = 2-5 m f , v n mT „ D cos a 160 X 0-9397 „ . - from (9) D = , T = = =-= = 60-13 cos a m 2-5 Since the ratio is to be 2: 1 let us try 60T and 30T, and find a new value for a. D cos a Then for the wheel from T = m mT cos a ■ = 2-5 X 60 150 ~ 160 ~ 160 = 0-9375 hence a = 20°22' To obtain the leads of the helices of the wheels we have for the wheel: Pitch cir cum = 160tt = 502-7 mm 502-7 502-7 and from (19) Lead = tan 2Qo22 , = ^^ = 1354mm For the pinion Lead = - — - Oo 'y = half of lead of wheel = 677 mm HELICAL GEARS 101 The particulars of the gears are thus as follows: Wheel No. of teeth = 60 Pinion. No. of teeth = 30 Pitch dia = 160 mm Pitch dia = 80 mm Add = 2- 5 mm Top dia = 85 mm Top dia = 165 mm Cutting depth = 5-625 mm Cutting depth = 2-25 m Helix angle = 20°22'. = 5-625 mm LH helix Helix angle = 20°22'. Helix lead = 677 mm RH helix. ~ ., + 30 Helix lead = 1 354m Cutter t0 USe= T053tSJ> _ 60 = cutter as for 36T spur Cuttertouse = (093W wheel. = cutter as for 73T spur wheel. The reader will observe that the solution to these problems is arrived at by a compromise after a system of trial and error. With wheels of this type such a procedure is nearly always necessary before a practical set of working conditions can be arrived at. Generally, the conditions allow one or more of the gear elements to be varied to suit the problem. Exercises 4b 1. Calculate the pitch diameter, top diameter and cutting depth for a 42T gear having J stub teeth. 2. Find the gear-tooth caliper settings for checking the tooth of a 32T, J stub-tooth gear. 3. A pair of gears are required to connect two shafts at 160 mm centres. Ifthe speed ratio required is 3:5, and the gears are to have £ stub teeth, find their leading particulars. 4. A pair of gears consists of 27T and 63T gears of 4mm module, 20° pressure angle. If the backlash allowance is 0-2 mm, what should be the centre distance between the two gears when the teeth are hard in contact? 5. A train of gears consists of the following: 30T driving 48T driving 48T driving 7 5T. The teeth are 4 mm module and the backlash allowance on all the teeth is 0-1 5 mm. If all the backlash is taken up in one direction, through what angle must the 30T gear be turned before the drive is taken up by the 75T wheel? 6. Calculate the following particulars for a 52T spiral gear of 4 mm normal module, 20° spiral angle: (a) pitch diameter, (b) top diameter, (c) cutting depth, (d) lead of spiral, (e) suitable cutter to use. 7. A helical gear is to have a helix angle of30°(RH), and the normal module is 4 mm. The pitch diameter must be as near as possible to 125 mm. Calculate (a) the number of teeth, (b) the pitch, and top diameters, (c) the lead of the helix, (d) the correct cutter to use. 102 CALCULATION FOR GEARS AND GEAR CUTTING 8. A spiral gear has 25 teeth, a helix angle of 45°, and an approximate pitch diameter of 75mm. Calculate the nearest normal metric module. Find also the pitch diameter, top diameter., lead of spiral and cutting depth. 9. A pair of spiral gears are required to connect two parallel shafts 100 mm apart, with a gear rat io of 3 : 2. Working on a normal module of 2-5mm, and an approximate helix angle of 20°, determine particulars of a suitable pair of wheels. Worm gearing A worn drive is often used to connect two non-interesting shafts which are at right-angles and a fair distance apart. The worm is the equivalent -Lead(L) Start Af?7 Starts N°. s 2&3 Start N92 Lead 'Angle (/{.) Fig. 79 Diagram of 3-Start Worm. of a screw thread, the shape of a section of the thread on a plane through the axis of the worm, being the same as a rack tooth in the involute system . The action between a worm and wormwheel is equivalent to that of the wheel as a gear; rolling along the worm as a rack. As the worm is usually produced by turning, or by a milling process similar to turning, the pitch Fig. 80 Worm Thread Form for 14£° Pressure Angle. WORM WHEEL 103 most commonly used is the circular pitch (p) . Fig . 79 shows the pitch, lead and lead angle for a worm, and Fig. 80 gives the proportions for the thread on a section through the axis for a tooth of 14|° pressure angle. The relationships between the pitch, lead and lead angle are the same as for a screw thread. Lead (L) = (Axial pitch)(No. of starts) = pn Lead L tan A = n (Pitch dia) nd The pitch diameter of a worm is quite an arbitrary dimension, and a worm may be cut to any pitch diameter suitable to accommodate it to centre distance at which it is to engage with the wheel. In general, the pitch diameter should not be less than four times the pitch. Worms with large lead angles The efficiency of a worm drive increases with the lead angle up to a maxi- mum at about 45° . In view of this, multi-start worms with large lead angles are to be preferred. Unfortunately, with such worms, interference dif- ficulties occur in cutting and operation, and when the lead angle exceeds 20° it is usual to increase the pressure angle (i.e. the included angle of the worm thread is increased). The pressure angle may be taken up to 20° (40° incl. angle of thread), and in very quick start worms iji is sometimes made 30° . For pressure angles other than 14£° the tooth proportions must be re-calculated on the basis of the new angle and will not be the same as those shown on Fig. 80. Also, when the lead angle exceeds 15°; it is more advantageous to base the tooth proportions on the normal pitch. For the case of the tooth shown in Fig. 80 these are modified as follows: Normal pitch = (axial pitch)(cos of lead angle) p n = p cos X (as (7) above) Then Addendum of thread = 0-318/»„ Depth of thread = 0-6866/?,, Width of finishing tool at bottom = 0-3 \p n Worm wheel For the best results the cross-section of the worm-wheel rim should be of the form shown in Fig. 81 (a). For light duty and moderate speeds, wheels as shown at 81 (b) will give satisfactory results. Teeth with curved bot- 104 CALCULATION FOR GEARS AND GEAR CUTTING Wor m Cen tre Wheel Axis (CL) mm Fig. 81 toms should be cut with a hob or a fly cutter; that at 81 (c) is really a helical gear and may be cut as such. The shape of the teeth of worm wheels is the same as for involute gears of the same pressure angle. Worm wheel dimensions These will be based on the circular pitch and for large worm lead angles the tooth proportions will be in terms of the normal pitch. In reading the following, Fig. 81 (a) should be referred to: Pitch dia (D) = Pitch circum. IE 71 Throat dia = Pitch dia + 2 add - D + 0-63/) Centre distance (C) = Pitch rad wheel + pitch rad worm _D d ~ 2 + 2 Throat rad (R t ) = C - \ throat dia Whole dia = 2(C- OA) = 2[C-*,cos(i0)] Width (w) approx = 2BC = 2 (Top rad of worm) [sin (£/J)] = (d + 0-636/?) sin (i/J) In cases where tooth proportions are based on normal pitch, then in expressions (19) and (23) above, p n should be used instead of/?. SPEED RATIO WITH WORM GEARING 105 Speed ratio with worm gearing If the wheel has T teeth, and the worm is single threaded (1 start), then the worm will turn T times for 1 turn of the wheel. A 2-start worm will turn T ■yr times per revolution of the wheel, and so on. Hence for T teeth in the wheel and n starts on the worm: Rev of worm T Speed ratio = 1 Rev of wheel n Example 7 . Determine the dimensions of a worm and wheel to operate at 120 mm centres and give a ratio of 16 — 1. Circular pitch 10 mm, pres- sure angle 14£°, wheel face angle 75°. For 10 mm pitch the minimum dia. of the worm should be 4 x 10 mm = 40 mm. This leaves 120 — 20 = 100 mm as the radius of the wheel. Dia of wheel = 200 mm. Circum = 200tt = 628-4 mm. At 10mm circular pitch this gives 63 teeth. If we make the worm 64T and use a 4-start worm, we shall obtain the 16—1 ratio required. Pitch dia wormwheel = = = 203-7 mm 71 71 203-7 and pitch rad = — ~ — = 101-85 mm Throat dia = 203-7 + 0-636/) = 203-7 + 0-636(10) = 21006 mm Throat rad = 120 - K2 10-06) = 1497 mm Whole dia = 2tl20 - 14-97 cos 37±°] = 2 [120 - 14-97 x 0-7949 = 216-24 mm Pitch rad worm = centre distance — rad wheel = 120 - 10185 = 18-15 and pitch dia = 2 x 18-15 = 36-30 mm Top dia worm = 36-30 + 0-636/? = 36-30 + 6-36 = 42-66 mm We may now find the approx. width of the wheel (h>) w = (d + 0-636» sin 37±° = 42-66 sin 37£° = 42-66 x 0-6088 == 25-97 mm (say 26 mm) tan of worm lead angle = — -r- = ^t^k = 0-3508 ° Ttd 7i x 36-30 X = 19° 20' Whole depth of tooth = 0-6866/> = 6.87 mm Width of threading tool at end = 0-31/? = 3- 10 mm 106 CALCULATION FOR GEARS AND GEAR CUTTING This gives us all the data necessary for the worm and wheel. As an exercise, the reader should make a full-size drawing of the pair. Helical (spiral) gears to mesh with worms Sometimes, as a compromise, and to avoid the delay and expense of ordering a new wormwheel, an ordinary straight-faced helical wheel is cut to replace a wormwheel. The main problem is to find, out of stock, a suitable cutter to give a satisfactory match-up. If stocks of circular, diametral and module pitch cutters are available, it is often possible to arrive at a workable solution. Let us see what could have been done to match a helical wheel to the worm in Example 7 above: Normal circular pitch of wormwheel = 10 cos 19°20' = 10(0-9436) = 9-436 mm 9-436 Normal module = = 3-003 71 This is very near a 3 mm module and a 3 mm module cutter would probably cut a satisfactory gear. Bevel gears Bevel gears are used to connect two shafts whose axes meet, and which are in the same plane. We saw in connection with spur gears that the motion of two gears was equivalent to that of two thin cylinders or discs rolling together, the diameters of the discs being the same as the pitch diameters of the gears. In the case of bevel gearing the fundamental conception of the motion is that of two cones rolling together. Fig. 82 BEVEL GEAR CALCULATIONS 107 In Fig. 82, OA and OB represent the axes of two shafts intersecting at O. COD and DOE are the two elemental cones, having OA and OB as their axes. The cones touch along the line OD, and if one is turned it will drive the other. In practice, the gears only consist of a narrow frustrum of the cones and are shown thickened. Metal is added at the back, as shown, to strengthen up the teeth in that region. The two elemental cones are called the pitch cones and become the imaginary pitch surfaces of the gears. The angle 6 P is the pitch angle of the pinion, W that of the wheel and I is the shaft angle. When teeth are cut in the gears, the number of teeth in each gear will be proportional to the pitch diameters CD and DE. Hence if n = rev/min of wheel and N = rev/min of pinion n T DE sin W N t CD sine, Note that 6 P + W =Z Bevel gear calculations /\ In Fig. 83: AOB = 6 is the pitch angle. AC = pitch dia (/)). /\ DOA = a is the addendum angle S\ AOF = /5 is the dedendum angle /\ EOB = $ = 6 + a is the face angle /\ FOB = p = 6 — ji is the root angle C is the cone distance. L is the tip distance. C /is the face width and may be made about = -~ D, L and C are given capital or small letters according to whether they refer to the wheel or to the pinion. The angles are generally given a suffix p or w to differentiate them, [p for pinion and w for wheel.] The back face DAF is always made perpendicular to the pitch surface AO, and the size and shape of the teeth as developed round that face correspond to the proportions for the pitch of the teeth in the gear 108 CALCULATION FOR GEARS AND GEAR CUTTING Add m Fig. 83 (e.g. if the gear had teeth of 5 mm module, then the teeth at the surface DF (as shown developed round line CH) would be the correct size for that pitch). In travelling down the tooth from face DF towards O, every line on the tooth converges to O . From Fig. 83: -=- -- C = sin 6 and C = -^—. — — 2 2 sin 6 tan a tan = Add m , . j i x — pr- = -^ (where m = module) Ded _ l-25m C ~ C Whole diameter (over corners DE) = pitch dia + 2AD cos 6 But AD = Add = m Whole dia = D + 2m cos Examples 8. Two shafts at 90° are to be geared together by bevels. The speed ratio is to be 3:2 and the pinion pitch diameter is to be 120mm. Determine the dimensions of 5 mm bevels. Since the ratio is 3:2 and d = 120mm D = 120 x I = 180mm T = 36 and t = 24 BEVEL GEAR CALCULATIONS 109 Fig. 84 Signifying the pitch angles by 6 W and p . Then tan P = ^ (Fig . 8 3). =% = \ 6 p = 33° 41' and B w Add = m = 5 mm Ded 90 - 33° 41' = 56° 19' 1-25/n = 6-25 mm OA = V90 2 + 60 2 = 108- 10 mm 5 tan of add angle tan ded angle 108-1 6-25 108-1 = 0-0463 a = 2° 39' = 0-0578 /$ = 3° 18' Face angle wheel <t> w = 56° 19' + 2°39' = 58° 58' Face angle pinion <j> P = 33°41' + 2° 39' = 36° 20' Root angle wheel P w = 56° 19' - 3° 18' = 53° 1' Root angle pinion P p = 33° 41' - 3° 18' = 30° 23' Whole dia (wheel) = D + 2m cos W = 180 + 10(0-5155) = 185- 16 mm Whole dia (pinion) = d + 2m cos P = 120 + 10(0-8056) = 128-06 mm Face width (if made }C) / = 120 20 6 x 0-5155 0-5155 A sketch of these wheels is shown at Fig. 84 '2 sin = 38-80 mm 110 CALCULATION FOR GEARS AND GEAR CUTTING Shafts not inclined at 90° When the shafts are inclined at angles other than 90° the calculation for the pitch angles is slightly more difficult. The following example will indicate a method of evaluating such cases. Example 9. Determine the dimensions of 4 module bevels to connect two shafts at 70°. The pitch diameter of the pinion is to be 80 mm and the ratio 4: 5. From the information given we have d = 80 mm D = 80 X | = 100mm 4 4 Z, = 70° In triangles OAC and OAB (Fig. 85), AC = 50 and AB = 40 Also Hence 7T-T- = sin 6 W and -~-r- = sin 6 P = sin (70 c OA OA 50 . . OA = Sin ^ M ^=sin(70< K) 3*39' (1) (2) Fig. 85 BEVEL GEAR CALCULATIONS 1 1 1 40 sin (70 c > K) 50 sin 9 W = sin 70° cos o w - - cos 70° sin W sin e w 0-9397 cos o w 0-342 sin W sin o w sin W Divide (2) by (1) ^ = sui u w sin 70° ens ft — ms 70° sin ft (see Appendix VI) 0-937 -0-342 tan W 4 0-937 ^ + 0-342 = f^4r- from which e w = 30° 22' 5 tan W w and p = 70° - Ow = 70° - 39° 22' = 30° 38' nA AB 4° 1QAQ OA = sin 30° 38' = 0^095 = 7848mm Add = module = 4 mm Ded = 4 x 1-25 = 5 mm 4 tan add angle = ^-^ = 0-05107 a = 2° 55' /o-4o tan ded angle = ^^ = 0-0638 /J = 3° 39' Face angle wheel W = 39° 22' + 2° 55' = 42° 17' Face angle pinion p = 30° 38' + 2° 55' = 33° 33'. Whole dia wheel = D + 2m cos W = 100+8 cos 39° 33' = 106- 18 mm Whole dia pinion = D + 2m cos P = 80 + 8 cos 30° 38' = 86-88 mm The main dimensions are shown on Fig. 85. Exercises 4c 1. Calculate the top diameter, root diameter and helix angle for a 2-start worm of 10mm pitch and 40mm pitch diameter. 2. If the worm in Question 1 drives a wheel, and the gear ratio is 17£ to 1, calculate the centre distance, and full particulars for the worm wheel. 3. A worm is 60mm pitch diameter, and 20mm circular pitch. Find the number of starts in order that the lead angle may be approximately 30°, and state the actual lead angle. Calculate the normal thickness of the thread on the pitch line. 4. A worm drive operates at 150 mm centres and the ratio required is 15 to 1. Taking a circular pitch of 10mm and a worm about 50mm pitch diameter, find a suitable worm and wheel for the drive. 5. Two shafts at 90° are to be connected by equal bevel wheels. Determine the dimensions of one of these if the module is 2-5 mm, and the pitch diameter 60mm. 1 12 CA LCULATION FOR GEARS AND GEAR CUTTING 6. Two shafts at 90° are to be connected by bevels to give a ratio of 3:2. If the pitch diameter of the smaller bevel is 130 mm and the module is 5 mm, determine the dimensions of the wheels. 7. A bevel gear has 20 teeth of 4mm module and a pitch angle of 45°. Determine the top diameter and included angle of a taper pin, which will rest in a tooth space with its centre on the pitch cone, its large end level with the back face of the gear, and its curved surface making contact with the tooth sides for its whole length. (ij> = 20°). 8. Two shafts, inclined at 120°, are to be connected by bevel wheels to give a ratio of 4:3. The pitch diameter of the smaller gear is 90mm and the teeth are 5mm module. Determine particulars of the gears. 5 Milling and the milling machine Milling Cutters For the purpose of considering the calculations necessary in connection with milling cutters, we may divide them into three general types: (a) those with fluted teeth, (b) machine relieved, (c) inserted teeth. Sketches of these are shown in Fig. 86. (a) fluted Cutter (b)Machine Relieved Cutter Fig. 86 (c) Inserted Blade Cutter Number of teeth Milling cutters and milling conditions vary so widely that it is difficult to set hard and fast rules for determining the number of teeth to be put in a cutter. For fluted and relieved cutters the rule N = 2-75 VD - 5-8 gives a reasonably proportioned tooth [N = No. of teeth, D = Diameter of cutter] . The formula N = -pr + 8 gives a fairly coarse tooth for cutters over 60 mm diameter. Take a 100 mm cutter, the first formula gives: AT = 2.75\/lOO - 5-8 = 22 teeth whilst the second expression gives N = }W + 8 = 16teeth- 114 MILLING AND THE MILLING MACHINE For an inserted-blade-face mill it is better to assess the number of teeth on the assumption of their being spaced a suitable distance apart on the periphery of the cutter. Thus if we take a 200mm face mill and assume the blades to be spaced 30 mm apart, we have No of blades = Circumference of cutter 200;* 30 30 = approx20 Rake on cutter teeth If a cutter has its teeth milled with their faces radial, and parallel to the cutter axis as shown in Fig. 87(a), the teeth have neither top nor side rake . If, however, the end view of the tooth is as shown in Fig. 87(6) the front rake is the angle a, since the tooth cuts relative to a radial line OA from the centre to its tip. Side rake is put on the tooth by milling it on a helix as shown at Fig. 87(c). The side rake is then equal to the helix angle /3. When teeth are of this form the relation between the hand of the helix and the direction of rotation of the cutter is important, in order that the end thrust introduced may be accommodated efficiently. (CL) No Top Rake No Side Rake (c) Side Rake Fig. 87. The calculation of the helix angle is discussed in the section on spiral milling. Top rake can be put on the teeth by milling it off-set as shown in Fig. 88. Instead of the front of the tooth being milled on the centre line it is RAKE ON CUTTER TEETH 115 cut off centre by distance x. Then ifR = radius of cutter and a the rake x angle required we see that — = sin a and x = R sin a . R Rakes up to 10° are advantageous, but above that angle they may cause the cutter to chatter. Blank being Fluted Fig. 88 Fig. 89 Example 1 . Calculate the amount of offset to give 10° of rake on the tooth of a cutter 80 mm diameter. In this case R = 40 and since sin 10° = 0-1736 x = 40 sin 10° = 40 x 0-1736 = 6-944 (say 7 mm) The case for an inserted-blade cutter is shown in Fig. 89, where ifR = radius over blades, then x = R sin a as before. Angle of fluting cutter For milling the flutes in cutters the problem arises of determining the angle a of the fluting cutter to give the required depth (d) of the flute (Fig. 90). 116 MILLING AND THE MILLING MACHINE Fig. 90 In the following analysis any land on the tooth is neglected, A being assumed as a sharp edge. Also, the point B is assumed as a sharp corner, so that if there is any radius on the cutter, B is the point at which the cutter edges CB and AB would intersect. The variation due to the first assumption tends to cancel out variations due to the second. AD is drawn perpendicular to CBO, then /\ AD = AO sin AOC = AD sin -rr- [where N = No. of teeth in cutter] and since AO = rad of cutter, R, 360 AD = R sin # Now tan a = R sin 360 AD AD AD " J1 " AT DB ~ CB - CD " d - CD ~ d - CD But CD = CO - DO = R - AO cos 1 — cos _ _ 360 J. 360\ R — R cos —fi- = R II — cos —fi- ] 360 360 ' N R sin Hence tan a = 360 </-*(. cos**) CUTTER CLEARANCE 117 Example 2. Calculate the angle of fluting cutter required to mill 16 teeth, 6 mm deep in a cutter 80 mm diameter. Here R = 40, N = 16 and d = 6 R sin tan a = 360 N 40 sin 22^° 6 -40(1 15-31 cos 22i°) 15-31 6 79°6' 3-044 2-956 = 5-18 360 N 360 16 = 22^ From which a The nearest cutter to use would probably then be an 80° cutter. Clearance The cutting clearance is put on the teeth at the time they are sharpened. The cutting edges may be ground either on the periphery of a disc wheel as shown at Fig. 91(a) or on the face of a cup wheel as at (b). When the teeth are ground on the periphery of a disc wheel as at (a), the centre of the cutter is set below the centre of the wheel and the radial Grinding Wheel Fig. 91 118 MILLING AND THE MILLING MACHINE line joining the tooth edge to the cutter centre is horizontal. The clearance angle obtained is shown as C, and in triangle OBA: AB . „ OB = SmC But OB = radius of wheel (R) and AB = offset (h c ) Hence -^- = sin C h c = R sin C Using the face of a cup wheel as at (b), the position of the cutter centre relative to the wheel is immaterial, but the tooth is set below the cutter centre by the distance h t and in triangle OED: ED = h t ; OD = rad of cutter (/•) ED . _ h t . _ t^ft = sin C or — = sin C OD r and h, = r sin C Example 3. Calculate the settings for grinding the teeth of an 80 mm spiral mill, (a) Using the periphery of a 200mm disc wheel and (b) using the face of a cup wheel. (Clearance required = 6°.) The first condition is as shown at Fig. 91(a) and h c = R sin 6° R = rad of wheel = 100mm :.h c = 100 sin 6° = 100 x 0-1045 = 10-45 mm Using the cup wheel we have as in Fig. 91 (b) h t = r sin 6° = 40 x 0-1045 = 4-18mm Effect of the helical tooth on the clearance When the tooth of the cutter is helical (spiral) the calculation for the above cutter setting is modified as follows: In Fig. 92 AK is perpendicular to the end of the cutter (parallel to its axis), so that /$ is the helix angle of the tooth. The clearance face is AC and /\ BAC = C a is the clearance angle referred to the end of the cutter, or to a plane perpendicular to the cutter axis. EFFECT OF THE HELICAL TOOTH ON THE CLEARANCE 1 19 Fig. 92 DG is parallel to AB and DF to AC, so that FDG = C a as before. DL is horizontal and perpendicular to the tooth face AD, and DE is also perpendicular to AD, but E is a point on the bottom of the clearance face. G and H are vertically above F and E respectively. y\ Hence EDH is the clearance on the tooth referred to a plane perpen- dicular to its front. Call this normal clearance (C„). The axial clearance (C a ) is that put on by the grinding wheel, whilst the normal clearance (C„) is the one effective when the cutter is in action. It is necessary, therefore, to grind such an axial clearance as will give us the required normal clearance, and for this purpose we require a relation between C a and C n . This may be obtained by considering triangles EHD, HDG and FGD, which are right-angled at H, H and G respectively. tan C„ = FG EH DG ~ DG (since EH = FG) But DH DG Hence from above: tan C, — cos /J and DG EH DG DH cos/J EH EH ' DH_=DH COS/} cos fl 120 MILLING AND THE MILLING MACHINE But — = tan C n Hence tan C a = tan C„ cos /$ Thus when we know the value required for the normal clearance (C„) and the helix angle /}, we can find the axial clearance angle C a for the grinding setting. Example 4. If in the last Example the normal clearance required was 6° and the cutter had a helix angle of 30°, calculate the axial clearance for setting. We have that tan C a = tan C n cos /$ C„ = 6° and tan C n = 0-1051 fl = 30° and cos = 0-866 tan C a = 0-1051 x 0-866 = 0-091 From which C a = 5° 12' For small helix angles the correction is not important, but it should be carried out on cutters with steep angles. With the machine relieved cutter the clearance is put on at the time the tooth is form relieved and the tooth is sharpened by grinding its front with a saucer-shaped wheel. When this is carried out care should be exercised to ensure that if the front of the tooth was originally radial (on the centre), this position is preserved for it. If this is not done, the effect on the formed profile will be similar to that we discussed when considering the circular form tools on page 73, and the accuracy of the cutter form will be lost. Exercises 5a 1. Calculate the following for an 80 mm diameter fluted cutter: (a) a suitable number of teeth to give a coarse pitch, (b) the off-set of the tooth front necessary to give 10° of front cutting rake, (c) the setting for grinding the teeth on a cup wheel, to give 7° of clearance. 2. 300mm face mill is to be fitted with blades 6mm thick. Estimate a suitable number of blades, and calculate the off-set of the blade necessary to effect 12° of top cutting rake. 3. The flutes of a 25 mm end mill are cut LH helix 500mm lead. Assuming the flutes to be 4 mm deep, calculate the helix angle, based on the mean diameter of the flutes. If the end teeth follow this angle, is the rake on them positive or negative? 4. A spiral cutter is 60mm diameter and the teeth are cut on a helix of 300mm lead. If an apparent clearance of 8° is ground on these flutes, what is the true clearance? SPEEDS AND FEEDS FOR MILLING CUTTERS 121 5. Calculate the nearest angle of cutter to use for fluting a 60 mm diameter cutter with 12 flutes, 6mm deep. 6. What depth of flute will be obtained by cutting 18 teeth in a 80 mm cutter with an 80° fluting cutter? Speeds and feeds for milling cutters The cutting speed for milling is found in the same way as for turning and drilling, so that if Q mm = diameter of cutter, and N rev/min its speed, Cutting speed (m/min) TtDN , .. 10005 = T006 andAr = ^zJ- Cutting speeds should be as high as possible consistent with an econo- mic cutter life before it needs re-grinding, and the speeds given for turning form a reasonable basis upon which to set the speed of a milling cutter. The rate at which the work feeds beneath a milling cutter is sometimes expressed in millimetres per minute and sometimes in millimetres per revolution of the cutter. Neither of these methods gives a reliable indica- tion of the cutter performance since both ignore the number of teeth in the cutter. The most equitable method of assessing milling feeds is in millimetres per tooth, since this gives an indication of the work each tooth is doing. From the feed per tooth, feed per revolution can be found by multiplying by the number of teeth, and a further multiplication by the rev/min gives feed per minute. The following table gives an indication of the feed per tooth possible with various types of milling cutters: Table giving feeds for H.S.S. cutters. Cutter Feed per Tooth mm Spiral (slab) mill (up to 30° helix angle of tooth) 01 to 0-25 Spiral mill (30°-60° helix angle) 005 to 0-2 Face mill and shell end mill 0-1 to 0-5 End mill 0-1 to 0-25 Saw 005 to 0-1 Slotting cutter 005 to 01 5 Form cutters 005 to 0-2 Example 5. Calculate a suitable speed and feed for a 80 mm spiral mill with 18 teeth to take roughing cuts on mild steel. 122 MILLING AND THE MILLING MACHINE For roughing cuts we may take a moderately slow speed with a heavy feed. Assume a speed of 20m/min and a feed of 0-2 mm per tooth. A , 10005 1000 x 22 x 7 __ , . N = -nD~ = 80x22 = 87 ' 5rev / min A feed of 0-2 mm per tooth gives 0-2 x 18 = 3-6 mm per rev. = 3-6 X 87-5 - 315 mm/min. Power absorbed in milling The conditions in milling are so variable that it is possible to attempt only a very rough computation of power requirements. From experimental work it has been found that the power requirements for milling are approximately as given in the table below. The figures should betaken only as a rough guide since the power varies with the amount of cut, the cutter, the cutting lubricant and other factors. However, even if the reader's, results are not all they might be, the working out will provide useful mathematical practice. ENERGY REQUIRED FOR MILLING* Values given are Joules per cubic millimetre removed Cast Mild Hard Alumin- Material being Cut ^ ^ ^ ^ mm J/mm 3 1-9- 2-7 4-0 to 7-0 1-60 0-90 *For face milling the power may be taken as f to \ of that given in table. Using the values given in the table, the energy being absorbed is found by multiplying the tabulated energy by the volume of metal being re- moved per minute. This is found by multiplying the depth of cut, the width of cut and the feed length. Thus if d = depth of cut; w = width of cut, and / = feed Volume = d.w.f., and Power = energy per second (watts) To allow for frictional losses in the machine add approximately 30% . THE DIVIDING HEAD 123 Example 6. A spiral milling cutter is taking a cut 4 mm deep with a feed of 120mm per min over a cast-iron block 80mm wide. Estimate the power required to drive the machine. Here the depth of cut (d) = 4 mm width of cut (w) = 80 mm feed = 120mm/min = 2mm/s Volume of metal removed per second = 4 x 80 + 2 = 640 mm 3 From the table we have l-9J/mm 3 for cast iron. .-. Power for cutting = 1-9 x 640J/s = 1216W = l-216kW Adding 30% for machine losses we have 1-216 X jjjjj = l-S8kW Exercises 5b 1. A 100 mm diameter spiral cutter has 18 teeth. Calculate the speed in rev/min and the feed in mm/min for this cutter to be operating at a cutting speed of 22m/min and a feed of 0- 1 5 mm per tooth per rev. 2. If a cutter in Question 1 was operating on a job 80mm wide, with a cut 4 mm deep, calculate the volume of metal removed per minute, and estimate the power input to the machine if the material being cut is cast iron and frictional losses are equivalent to 30% of the cutting power. 3. If a milling machine is equipped with a 4kW motor, estimate the deepest cut that may safely be taken on hard steel, when the work is 100 mm wide and the feed = 150 mm per min. (Take cutting power as 75% of motor rating.) 4. A face milling cutter is 300 mm diameter and has 28 teeth. If this is operating at a cutting speed of 33m/ min, and a feed of 0.2mm per tooth, how long will it take to travel over a cut 400 mm long? 5. A certain job can be milled either with a 80 mm spiral mill, or with a 150 mm face mill. Each cutter has 16 teeth and the cutting speed to be employed is 22m per min. If the feed for the spiral mill is 0-2 mm per tooth and for the face mill 0-25 mm per tooth, which is the most economical method of working? 6. A 25mm end mill with 8 teeth is milling a slot 10mm deep at a feed of0-025mm per tooth, and a speed of 600 rev/min. If the material being cut is brass, estimate the power input (assume 30% loss) and the time and power cost for milling a slot 300 mm long. Take power at 2p per B.O.T. Unit [1000 watt hours] . The dividing head The dividing head, which is shown diagrammatically'in Fig. 93 is used with the milling machine for the purpose of obtaining divisions of the 124 MULING AND THE MILLING MACHINE 40 T. Worm wheel End for attachmem of work Taper hole Single start worm Plunger for locking index plate Crank Fig. 93 circle. It consists essentially of the spindle to which is attached a 40-tooth wormwheel. Meshing with this is a single-threaded worm, to the spindle of which is attached the indexing crank. Adjacent to the indexing crank is the index plate containing several series of equally spaced holes arranged in circles on its face. A pin in the indexing crank can be adjusted so that its radius coincides with any of the hole circles, and an adjustable sector enables any proportion of the index plate circumference to be divided off. Since the gear ratio in the head is 40-1, 40 turns of the crank cause the spindle (and the work attached to it) to make 1 turn, or 1 turn of the crank rotates the spindle ^th of a turn. The object of the index plate with its holes is to subdivide further the turn of the crank, and the greater the range of hole circles available the greater will be the number of divisions possible without resource to special indexing methods. The Brown and Sharpe dividing head is provided with three indexing plates having hole circles as follows: Plate No. 1: 15, 16, 17, 18, 19, 20 holes. Plate No. 2: 21, 23, 27, 29, 31, 33 holes. Plate No. 3: 37, 39, 41, 43, 47, 49 holes. The standard Cincinnati dividing plate is of larger diameter than those used on the Brown and Sharpe head and is reversible. It is provided with the following hole circles: On one side: 24, 25, 28, 30, 34, 37, 38, 39, 41, 42, 43 holes. On the reverse side: 46, 47, 49, 51, 53, 54, 57, 58, 59, 62, 66 holes. SIMPLE INDEXING 125 This plate enables all divisions up to 60 to be obtained in addition to all even numbers, and numbers divisible by 5, up to 120. In addition to this standard plate, special ones can be obtained when divisions beyond its range have to be indexed. Simple indexing The majority of divisions required can be obtained without difficulty by indexing in one of the sets of hole circles supplied. Straightforward working in this way is usually termed simple indexing. Since 40 turns of the crank cause 1 turn of the work, if we require n equal divisions in the work each division will be -th of its circum- n 40 ference and the turns required of the crank will be — . It will help the reader to remember which way up this fraction should be if he remembers that when more than 40 divisions are required on the work the crank will have to rotate less than one complete turn. Example 7. Calculate suitable indexing for the following numbers of divisions: (a) 6, (b) 10, (c) 15, (d) 22, (e) 28, (J) 37, (g) 48, (h) 62. (a) 6 divisions. Indexing = f = 6£ = 6f = 6# or 6JJ = 6 whole turns + 14 holes in a 21 circle or 16 holes in a 24 circle or any combination giving f of a turn (b) 10 divisions. Indexing = fg = 4 complete turns (c) 15 divisions Indexing = $ = 2f& = 2f = 2 whole turns + f of a turn [see (a) above] (d) 22 divisions. Indexing = f§ = ljf = lft = lfi or Iff = 1 whole turn + 27 holes in a 33 circle or 54 holes in a 66 circle (e) 28 divisions. Indexing = ff = ltf = If = 1£ or l|f = 1 whole turn + 9 holes in a 21 circle or 18 holes in a 42 circle 126 MILLING AND THE MILLING MACHINE if) 37 divisions. Indexing = $ = 1£ = 1 whole turn + 3 holes in a 37 circle (g) 48 divisions. Indexing — iQ — i — il— 20 lllUCAUlg — 48 — 6 — lg — 24 = 15 holes in an 18 circle or 20 holes in a 24 circle (h) 62 divisions. Indexing = f§ = §? = 40 holes in a 62 circle or 20 holes in a 3 1 circle Compound indexing When a division is required which is beyond the capacity of the available hole circles, a method of compound indexing may be used. The index plate is usually locked and prevented from turning by means of a plunger fitting in one of the circles of holes. The principle of com- pound indexing is to obtain the required division in two stages. (1) By a movement with the crank in the usual way. (2) By adding or subtracting a further movement by rotating the index plate and controlled by the plate locking plunger. Suppose the crank is indexed 5 holes in a 20-hole circle and then the index plate, together with the crank, is indexed a further hole with the locking plunger registering in a 15-hole circle. If both movements have been made in the same direction the total indexing will have been to + rV = & + & = m on the worm. If the plate had been turned opposite to the crank we should have had J L — 11 _ J_ — - li 20 15 — ' 60 60 — 60- By compounding suitable hole circles in this way it is possible to obtain a large number of additional divisions. 40 If n is the number of divisions required on the work, then — is n the indexing required and the fractions representing the two movements 40 to be used must give — , either when added or subtracted. Also the n denominators of the two fractions must be numbers equal to available hole circles in the plate. Suitable hole circles must generally be deter- mined by a method of trial and error and the following examples will illustrate the method. COMPOUND INDEXING 127 Example 8 . Determine suitable compound indexing for the following divi- sions: (a) 77, (b) 91. (a) 11 divisions. The indexing required is $ and we require two suitable fractions which give this when added or subtracted. The method of trial and error may be assisted by the following work- ing. Put down the 77 above a line, the 40 below it, and factorise them. 77 = 11 X 7 40 = 2x2x2x5 The numbers representing hole circles are now required to be written below the 40 and factorised. Their difference, also factorised, must be written above the line. The numbers must be so chosen that all the factors above the line must cancel out with numbers below. [The reader should notice that since only one plate can be used, the numbers must be those of two hole circles on the same plate.] Choosing 21 and 33 as the numbers, we have 21 = 7 x 3; 33 = 3 x 11 and the difference 12 = 2 x 2 x 3. Putting these numbers with their factors down we find that all the numbers above the line will cancel thus 77 = 44 X 7- 12 = 2- x 2- x 3- 40=2-x2-x2x5 21 = 1- x 3- 33 = 3 x 44 Hence 2 1 and 33 circles will be suitable and we require to find the res- pective number of holes to be indexed. Let these be a and b. tu a ^ b 40 Then 21 ± 33 77 (7 X 3) (3 x 11) (11 X 7) Putting on a common denominator 11a ±lb = 3 x 40 7x3x11 i.e. 11a ±lb = 120 128 MILLING AND THE MILLING MACHINE By trial and error we find that if a = 9, and b = 3; 99 + 21 = 120. Hence the indexing required is 9 holes in a 21 circle added to 3 holes in a 33 circle (i.e. both movements in the same direction). (Jb) 91 divisions: Testing for suitable hole circles as before we have: 91 = 44 X ^ 10 = 4 x 2- 5- 40 = 2 x 2 x 2- X 39 = 44 x 3 49 = 7 x 2 49 as suitable sizes and a b 40 39 ± 49 = 91 (13 x 3) (7X7) (13 x 7) Putting on a common denominator 49a ± 39b = 40 X 21 13 x 7 x 7 x 3 i.e. 49a ± 39b = 840. By trial and error: If a = 6 and b = 14: 49 X 6 + 39 x 14 = 294 + 456 = 840. Hence 6 holes in a 39 circle added to 14 holes in 49 circle will be the indexing required. Differential indexing This is really an automatic method of carrying out compound indexing. The arrangement of the dividing head is shown in Fig. 94, from which it will be seen that the index plate is unlocked, and is geared back to the spindle. As the spindle is rotated via the crank and worm, the gear train causes the index plate to turn backwards or forwards, and the net result is the same as if the index plate were released and rotated by hand as in compound indexing. Differential indexing is more straightforward and is capable of dealing with a wider range of divisions than compound indexing. The problem is DIFFERENTIAL INDEXING 129 Index p/ate 'unlocked Gear fixed to index p/ate Worm shaft turned by crank Gear fixed to bevel wheel Bevel wheels and gears A+Bgive J/7 ratio Driver Gear stud fixei to head spindle Fig. 94 Showing Arrangement of Gearing for Differential Indexing. to calculate the indexing and the gear ratio necessary to obtain any given number of divisions on the work, and we will explain the method by work- ing out a few examples: Example 9. Calculate the differential indexing to give 107 divisions on the work. The indexing required is ^ turns of the crank per division, and when this indexing has been done 107 times the crank has turned ^ X 107 = 40 times, i.e. the work has turned 1 complete circle as it should. Since no 107 circle is available, let us take an approximately near indexing to the exact ^ required. ToV = i\ approximately (by approx cancellation by 5) If we take 107 moves of 8 holes in a 21 circle we obtain 107 x 8 21 _ 416 ~~ 21 40^ turns of the crank 130 MILLING AND THE MILLING MACHINE But we have seen that the crank must make 40 turns only, during the 107 indexings, and we must therefore subtract jf of a turn. This is done by gearing up the plate, so that whilst the spindle makes 1 turn, the plate makes -If turn in the opposite direction to the crank. Hence with an indexing of 8 holes in a 21 circle, the gear ratio from . ,, , , x Drivers 16 spindle to plate: -=-^ = ^rr ^ ^ Driven 21 For the Brown and Sharpe dividing head, the gears supplied are as follows: 24(2), 28, 32, 40, 44, 48, 56, 64, 72, 86 and 100 teeth. We may make up a train for the above ratio from this set as follows: 16 8 X 2 _ 64 32 Drivers 27 ~ 7~x~3 ~ 56 X 48 Driven In the above case the gears must be arranged so that the index plate revolves opposite to the crank. Example 10. Calculate the differential indexing for 127 divisions. Exact indexing = -$; and roughly cancelling by 8 gives -^ as the approximation. Now 127 moves of 5 holes in a 16 circle gives 127 x & = W = 39f^ turns of the crank. But this is 40-39{£ = re short of what it should be. Hence in 1 turn of the spindle the index plate must make ^ turn in the same direction as the crank. 4 . Drivers 5 5 x 1 40 24 Gear ratio: t=-^ = tz = o » *•> = Ta x 7q Driven 16 8x2 64 48 With this ratio and an indexing of 5 holes in a 16 circle, the 127 divisions would be obtained. Angular indexing Very often, instead of a number of equal divisions, an angle must be indexed . Since 1 turn ofthe crank rotates the spindle^ turn, the angle at the work 360 centre equivalent to one turn of the crank is -^- = 9° so that ANGULAR INDEXING 131 rx, i Angle required Turns of crank to give any angle = ~ Example 11. Calculate the indexing for the following angles: (a) 41°, (b) 15° 30', (c) 29° 20'. (a) 41°. Indexing = ^ = 4| turn of crank, say 4 whole turns and 30 holes in a 54 circle. (b) 15° 30'. Indexing = !|£ = 1*| = ift = 1 whole turn and 1 3 holes in an 18 circle or 39 holes in an 54 circle id) 29° 20'. Indexing = th. = i±L = 3^ = 3£ = 3 complete turns and 7 holes in a 27 circle or 14 holes in a 54 circle Exercises 5c 1. Calculate suitable indexing to obtain the following divisions on the Brown and Sharpe head: (a) 12, (b) 15, (c) 22, (d) 34, {e) 41, (/") 50, (#) 62, (h) 76 divisions. 2. Calculate suitable indexing on the Cincinnati head for the following: (a) 13, (b) 17, (c) 25, (</) 36, (e) 45, (/) 54, (g) 65, (A) 82 divisions. 3. Find suitable indexing for the following angles on the Brown and Sharpe head: (a) 15°, (b) 26°, (c) 33° id) 52° 30', (e) 63° 40'. 4. Determine appropriate indexing for the following angles on the Cincinnati head: (a) 16°30', (b) 27°45', (c) 31° 20', (^ 74° 15', (e) 136° 30'. 5. A shaft, 50 mm diameter, is to have a groove milled along it. The sides of the groove are radial, it is 1 1-25 mm wide at the top and 6mm at the bottom. The centre is to be cut with a cutter 6 mm wide, after which the shaft is to be indexed round and set over for milling the slot sides with the same cutter setting. Calculate the indexing and set over. 6. Determine suitable compound indexing for the following, using B. & S. plates: (a) 51, (b) 63, (c) 87, (d) 189 divisions. 7. The crank of a dividing head is indexed N holes in a C circle, and then the plate is indexed in the opposite direction « holes in ac circle. Find an expression for the number of divisions obtained. 8. Determine suitable indexing and gears for obtaining the following by differential indexing: (a) 97, (b) 53, (c) 101, (d) 131 divisions. [Use B. & S. plates and gears.] . , . „,..,.,.. . . ,, . , . turns of spindle 9. The index plate of a dividing head is geared to the spindle in the ratio — . . — = 2 j (rotation opposite to crank) . If the crank is now indexed 3 complete turns, + 15 holes in a 20 circle, through what angle has the spindle been rotated? 132 MILLING AND THE MILLING MACHINE 10. A round plate requires four notches, A, B, C and D, indexing in it spaced so that the angles between AB, BC, CD and DA are in the ratio 2:3:4:5. Using the same hole circle throughout, determine the indexing required. Use of continued fractions for angular indexing When an angle is required whose value involves obscure minutes and seconds, it is unlikely that an exact indexing for it will be possible. It is possible, however, by turning the ratio into a continued fraction, to obtain an indexing which is very near to the exact one. The method is illustrated by the following examples: Example 12. Calculate the nearest indexing and the actual angle obtained for the following: (a) 14° 38', (6) 21° 19' 35" (a) The indexing will be — g — = 1— ^ — Converting the fraction to minutes gives lffjj = l^fg. We now convert |f| to a continued fraction and find its convergents. 169)270(1 169 101)169(1 101 68)101(1 68 33)68(2 66 2)33(16 32 1)2(2 The fraction is 1 +1 1 + I 1 + 1 2 + 1 16 + USE OF CONTINUED FRACTIONS FOR ANGULAR INDEXING 133 The convergents are as follows: 1st = f; 2nd = ±; 3rd = f; 4th = f; 5th = $■; 6th = \ 69 270 The 4th convergent is the last one we are able to make use of and i = 10 flr 15 8 16 Ul 24- Hence the indexing is 1 complete turn and 10 holes in a 16 circle or 15 holes in a 24 circle. The actual angle obtained will be 9 x If = 14f° = 14° 37*' (an error of -*')• 21° 19' 35" 3° 19' 35' (b) The indexing in this case will be s = 2 s and con- verting the fraction to seconds = 2£%$ = 2^ We now convert the 4& to a continued fraction 479 T296" 479)1296(2 958 338)479(1 338 141)338(2 282 56)141(2 112 29)56(1 29 The fraction is 27)29(1 27 2)27(13 26 02(2 2+ J 1 + 1 2 + \ 2+ I 1+J 1 + \ 13 + * 134 MILLING AND THE MILLING MACHINE and its convergents are 1st = \\ 2nd = \\ 3rd = f; 4th = ^; 5th = $; 6th — a- 7th — ia±- 8th — ^^~ Ulll — 46 , /III — 625 , OU1 — !296- If a 46-hole circle is available the 6th convergent may be used and the indexing will be 2 whole turns +17 holes in a 46 circle, giving an angle of 9° x 2# = 21±f° = 21° 19'33" (an error of -2"). If a 46-hole circle cannot be used, then the 5th convergent must be in- dexed and this will be 2 whole turns +10 holes in a 27 circle. This will give an angle of 9° x2ff= 21}° = 21° 20' (an error of + 25"). Spiral milling For some reason unknown, the operation of milling a helix on a cylindrical piece of work has come to be known as spiral milling. Actually a spiral is a fiat curve shaped like a clock spring. However, so long as we know the process and its principles, it matters little by what name we call it. Index plate unlocked^*. Driven Shaft i marked r 'C 'J In Fig. 94 1>J Idler F.g. 95 End View of Dividing Head Geared to Leadscrew for Spiral Milling. SPIRAL MILLING 135 When the machine is set up for spiral milling the worm spindle of the dividing head is geared to the leadscrew of the machine table, so that when the leadscrew is turned the worm is turned also. This rotates the dividing head spindle, so that the longitudinal movement of the table is accompanied by a rotation of the work. A sketch of the machine set up is shown in Fig. 95. The first calculation necessary for spiral milling is the gear ratio bet- ween the leadscrew and the dividing head worm-shaft to give the required lead of helix. In order to do this we must first ascertain the "lead of the machine." The reader will recollect that the lead of a screw is the distance the screw advances along the cylinder whilst it makes one complete turn round it (Fig . 78a) . The lead of the machine is the lead of the helix it would cut if the table leadscrew were connected to the dividing head worm by a f gear ratio. When this is the case we know that to rotate the dividing head spindle (and the work) through 1 revolution requires 40 turns of the worm. If the gear ratio to the table leadscrew is f the leadscrew will have made 40 turns also, and the table will have advanced 40 (pitch of leadscrew). This distance will be the lead of the machine. On the majority of milling machines having metric leadscrews the pitch is 5 mm, hence the lead of these machines is 200 mm (on machines manufactured to inch dimensions the leadscrew pitch is 0-25 in, hence the lead of these machines is 10 inches). When the lead of the machine is known, the gear ratio to cut a helix of any lead is given by the proportion: Drivers Lead of machine Ratio = Driven Lead of helix to be cut The hand of the helix (whether RH or LH) is controlled by the presence or otherwise of an idler gear in the train. Example 13. Calculate suitable trains of gears to cut helices of the follow- ing leads on a machine with a 5 mm leadscrew: (a) 120 mm lead, (b) 256mm lead, (c) 480mm, (d) 720mm lead. [Select from the B. & S. set of gears given on p. 130.] As the leadscrew is 5 mm pitch the lead of the machine will be 5 x 40 = 200 mm. 136 MILLING AND THE MILLING MACHINE (a) 120 mm Drivers Ratio j^t = ff = |fi and this is given by a 40 T driving a 24 T. The train may be compounded as follows: 40 5 X 8 _ 40 64 Drivers 24 ~ 4 X 6 ~ 32 X 48 Driven (b) 256mm lead. . Drivers _ 200 _ 100 x 2 100 x 24 Driven ~ 256 ~ 64 x 4 ~ 64 x 48 (c) 480mm lead Drivers 200 _ 40 _5_ 40 32 Driven ~ 480 ~ 48 X 10 ~ 48 X 64 (d) 720 mm lead _200_5_40_40 32 Ratio - 7 20 ~ 9 X 80 ~ 72 X 64 Helix angle Before the helix can be cut it is necessary to set the cutter to the angle fol- lowed by the path of the curve. If this were not done a great deal of inter- ference would take place and the shape of the groove would be nothing like the shape of the cutter working to produce it. Even with the cutter set into the helix angle some interference generally takes place and the shape of the groove varies from the profile of the cutter. The helix angle is the angle a on Fig. 78 (b), and if the helix is developed out into a triangle as shown, we see that circumference of cylinder _ lead of helix When the helix angle has been determined, the cutter head or the machine table must be swung round so that the plane of the cutter lies at this angle, relative to the work. Care should be exercised to swing in the correct direction for RH or LH helix. When grooves of appreciable depth are being cut, the circumference of the cylinder passing through the bottoms of the grooves will be much less than that of the cylinder in which they are being cut. This means that if the calculation for the helix angle is based on the outside diameter of the work, the inclination of the cutter will be correct at the top, but not at the bottom of the grooves. The reverse will be the case if the bottom HELIX ANGLE 137 diameter of the grooves is used when finding the cylinder circumference in the above expression. When in doubt about which diameter to take for the calculation the reader is advised to take the mean between top and bottom of grooves. It may be that for grooves of certain shapes the top, or the bottom, might form the more suitable basis for the calculation, but only trial and exper- ience can decide on the best compromise. Example 14. Calculate the gears and setting for milling LH spiral flutes in a reamer 40mm diameter. Lead of helix = 800mm. Machine leadscrew 5mm pitch. Reamer flutes 8mm deep. Lead of machine = 40 x 5 = 200 mm Gear ratio, table to head: r— - = ^r^ = t=o x t = 7o x ZT Driven 800 4 2 2 48 64 To calculate the helix angle we will take the diameter at the mean depth of the flutes, i.e. 40 — 8 = 32 mm. Circum of cylinder = 327rmm Circum 32;r n «,-«- tan a = — 1 — - r - = ^r = ^f = 0-1257 lead 800 25 From which a = 7° 10' As the helix is LH the table of the machine must be swung with its LH end away from the operator. Example 1 5 . A spiral gear has a pitch diameter of 80 mm and a spiral angle of 30°. Calculate the gears for milling the teeth in it. In this case we have to determine the lead of the helix in order to be able to solve the problem. The pitch circum = 80 n = 25 1-3 mm circum Also tan a = lead , , circum 251-3 251-3 A ~ c „ .-. lead = — = ~7^- =rrznHA = 435-2 mm tan a tan 30° 0-5774 • , _■ j- -j- i. j 200 1000 125 Gear ratio, leadscrew to dividing head = -tttj = yyfz = 972 138 MILLING AND THE MILLING MACHINE As this is an awkward fraction we will convert it to a continued frac- tion 125)272(2 250 22)125(5 110 15)22(1 11 7)15(2 ]4 1)7(7 7 The convergents are given by \ 2 + 1 5 + 1 1 + 1 2 + } and their values are: 1st = fc 2nd = A; 3rd = &; 4th #; 5th = §#• Taking the 3rd convergent and finding a suitable gear ratio we have Drivers 6 30 J_ _ 30 40 Driven ~ 13 ~ 65 X I ~ 65 X 40 The actual lead obtained will be ^ x 200 mm = 433-3 mm instead of the required 435-2 mm Cam milliing on the dividing head By employing a universal milling machine fitted with a swivelling vertical head, constant- rise cams may be cut on blanks held in the dividing head. The set-up is shown in Fig. 96, and the dividing head is geared to the table lead screw in the same way as for spiral milling. The principle of the operation is that as the table moves to the right the axis of the end mill approaches nearer to the axis of the dividing head spindle. If, at the same time, the dividing head is geared so that its spindle rotates, the combination of the rotation and of the cutter approaching CAM MILLING ON THE DIVIDING HEAD 139 ^Vertical head swung over Shaft marked C in Fig. 94 Gear connection similar to that for spiral milling (Fig.9S) Table lead screw Fig. 96 Set-up for Cam Milling. nearer to the centre, causes a cam to be cut on the blank held in the head. The rate at which the end mill approaches the dividing head axis is con- trolled by the angle a, and the reader will observe that if this is made zero (dividing head spindle horizontal), then, instead of a cam, a circular disc would be cut provided the end mill was long enough to do it. (It will be observed that as cutting proceeds, the blank gradually moves down the end mill so that a fairly long cutter is necessary.) (As an alternative to the arrangement shown, the end mill axis may be located below the cam axis. Then the table must move to the left and the cam will move up the end mill as cutting proceeds.) 140 MILLING AND THE MILLING MACHINE The expressions for calculating any particular case may be derived as follows: Let a == angle of inclination (Fig. 96) (Note that both cutter and dividing heads must be inclined at a) R =-- Gear ratio -=p between leadscrew and dividing head worm p =- pitch of table leadscrew / == lead of cam to be milled (Radial drop in profile in 1 revolution) Then, for 1 turn of the table leadscrew, the dividing head worm will turn R times and the dividing head spindle, because of the 40-1 reduction, D will turn 2^ times Hence, since in 1 turn of its leadscrew the table advances pmm we have: R_ Turns of work 40 R Movement of table p 40/) From which Turns of work = -r~- (movement of table) For 1 turn of the work: 1 = -Tpj— (movement of table) and movement of table = —J- (for 1 turn of work) K Now because of the inclination of the head and work, when the table has moved the distance cb (Fig. 97) the centre of the end mill has approach- ed the centre of the work by the length ac. [^-Movement of Tafa/e->-| Fig. 97 CAM MILLING ON THE DIVIDING HEAD 141 Hence if cb represents the horizontal table movement for 1 turn of the work, ac will represent the lead of cam cut and from the triangle: ac I -r = sin a or — t-. = sin a cb table movement Hence Table movement = sin a If we substitute this in the expression for the table movement above we have: Table movement = — =/-, i.e.—: = -=r- R sin a R , „ 40p sin a and R = — ^— lip = 5 mm, as is usually the case, the expression reduces to 200 sin a R = I This expression controls the variables R and a , and in using it a value for one of them will have to be assumed before the other may be calculat- ed. If a first trial gives unsatisfactory conditions, then one of the values may be changed to bring the other to reasonable dimensions. Example 16. Calculate a suitable setting for milling a cam, the profile of which falls 12 mm in 100° of its angle. Leadscrew of machine has a pitch of 5mm. Here p = 5 mm, and if the cam profile falls 1 2 mm in 100° , its lead will be 12 x -tqtt = 43-2mm Let us assume that R = 3 -,, „ 40o sin a ~ 200 sin a Then R =—i : 3 = 43-2 3 x 43-2 0-648 = sin a This gives a = 40° 24' Drivers 3 Hence a gear ratio -=r~- = y and an angular setting of a = 40° 24' will be suitable. 142 MILLING AND THE MILLING MACHINE Exercises 5d 1. Calculate the nearest indexing for the following angles, using the standard Cincinnati plate, and determine the actual angle obtained for the indexing used: (a) 10° 36' 30", (b) 41° 24'20", (c) 75° 45'30" . 2. Determine the nearest indexing to divide an angle of 76° 30' into 7 parts. Use hole circles available on B. & S. plates. 3. A plate is rectangular, 123 mm x 79 mm, with a hole in its centre. Ifthis is on a mandril between dividing head centres, find the nearest indexing to rotate it through the angle con- tained by joining the end corners to the centre. [Angle contained by short side required.] In the following exercises use the B. & S. gears (p. 130) and take the Lead of Machine as 200 mm. 4. Calculate suitable gear ratios to cut the following leads on a B. & S. machine: (a) 168 mm, (b) 288 mm, (c) 450 mm, (d) 640 mm. 5. A gear has a pitch diameter of 100mm and the lead of the spiral is 420mm. Determine a suitable gear ratio and find the helix angle for setting. 6. Calculate the gear ratio and angular setting for milling a 4-start thread of35 mm pitch, on a worm of 100mm pitch diameter. 7. A spiral milling cutter is 80mm diameter and 120mm long. Flutes are to be milled on it which make approx £ of their lead in the length of the cutter. Calculate suitable gears and angular setting for milling the flutes. 8. Slots h aving a lead angle of 30° are to be milled in a cylinder 45 mm diameter. Determine the gear ratio. In the following examples take the table leadscrew pitch as 5mm. 9. Calculate the lead of a cam which rises 10 mm in 50° of revolution, and determine a suitable setting and gear ratio for cutting it. 10. A cam which revolves at 2 rev/min has to move the roller in contact with it at a rate of 50mm/min. Determine its lead, and find a suitable gear ratio and setting to mill its profile. 11. A cam is heart-shaped with uniform rise. The radius to the point of the heart is 60 mm, and to the corner where the curved portions meet the radius is 30 mm. Determine the lead and find a suitable gear ratio and setting to mill it. The calculation of solid angles Angular milling and shaping jobs often occur where a solid angle has to be calculated, the value of the angle depending upon the angles between other surfaces on the same work. Examples of such cases are likely to be varied, but the following worked cases may convey to the reader how any other example might be approached. Example 17. Calculate the true angle between the sloping faces of the block shown in Fig. 98. If we project a section through the angle, the section plane being taken at 90° to the sloping corner, we shall obtain the triangle ABD shown at the bottom oJ the diagram. The true angle we require will then be ADB. THE CALCULATION OF SOLID ANGLES 143 Fig. 98 Now DE = CE cos 30° and since triangle ABC is right angled at C and has 45° angles at A andB Hence CE = ±AB = 5. DE=5cos30° =4-33 mm AE /\ In triangle ADE (sectional view): ^^ = tan ADE 5 ^N = tan ADE = 1-155 4-33 /\ From which ADE = 49° 7' /\ /\ ADB which we require is twice ADE = 2 x 49° 7' = 98° 14' Example 18. Calculate the angle between the base and the sloping face of the block shown in Fig. 99. The block is shown by full and plain dotted lines. Chain dotted lines are constructional only, for the purpose of explanation. If AB is perpendicular to AG and BE, and BC perpendicular to BE (BC being a line on the base), then the angle we require is angle ABC. FC is parallel to EB, and BD to FE 144 MILLING AND THE MILLING MACHINE Then in triangle BDC: BD = FE = Fig. 99 10 10 5-774 /\ tan 60° 1-732 BDC = 45° so that BC = BD sin 45° = 5-774 x 0-7071 = 4-083 10 AC /\ But-o^ == tan of ABC and AC /\ 10 .\ tan ABC = 4-083 ABC = 67° 49 2-453 Example 19. Calculate for the tool shown in Fig. 100. (a) The true angle between the cutting edge AB and the line AC (b) The inclination of the top cutting face, perpendicular to AB (c) The true angle between the top cutting face and the front clear- ance face. Lines FE on the cutting face and EG on the clearance face are perpen- dicular to the edge AB. FA is on the cutting face and is perpendicular to AC FH and EK are vertical lines and angles FHG and EKG = 90° In Fig. 100 (b), EM and FS are horizontal, LM vertical and EL is per- pendicular to AC. (a) True angle between AB and AC EL If we call this angle a, then -ry- = tan a EM EM Now ^rr- = cos 25° and EL = ^r- EL cos 25° THE CALCULATION OF SOLID ANGLES 145 Fig. 100 Also and and Hence Hence AL = Projected length AM EM EM Projected AM AL EM = tan 30° AL tan 30° EM tana = EL cos 25° AL EM EM tan 30° X cos 25° EM tan 30° tan 30° 0-5774 = 0-6371 cos 25° 0-9063 a = 32° 30 (b) Inclination of top face perpendicular to edge AB This will be the inclination of line FE 146 MILLING AND THE MILLING MACHINE Now since points A, F, E and L are all in the same plane, and FAL = FEA = 90°; AFE = EAL = a = 32° 30' FE = FA cos a and FA = OR = PQ = —^ cos 25° cos 25° Hence FF 21 — 21 cos 32° 30' 21x0-8434 ,__. FE == c^25^ C0S a = ~^Si2P = 0-9063 = 19 - 54mm The vertical distance of F below E = ES == Distance of F below A - Distance of E below A = PR - LM = PQ tan 25° - EL sin 25° =21 tan 25° - EL sin 25° == 21 x 0-4663 - EL sin 25° = 9-79 - EL sin 25° But EL = EA sin a and EA = FE tan a .'. EL = FE tan a sin a Hence EL sin 25° = FE tan a sin a sin 25° which, since FE = 19-54, and a = 32° 30' = 19-54 x 0-6371 x 0-05373 x 0-4226 = 2-83 PR - LM = 9-79 - 2-83 = 6-96 = ES. The inclination of the top cutting face in a plane perpendicular to AB = Angle EFS FS /\ 6.06 and H = sin EFS = -^- - 0-3563 FE 19-54 y\ From which EFS = 20° 52' (c) True angle between top face and front clearance face. This will be FES - 8° = 90° - 20° 52' - 8° = 90 - 28° 52' = 61° 8' Exercises 5e 1. Determine the angle, when measured perpendicular to the clearance face, of a tool for cutting acme threads (29° on its top face). Clearance angle on tool = 15°. 2. The angle of the vee in the block at Fig. 101 is to measure 60° on the front face as shown . Calculate the angle of the vee when measured along its slope (i.e . the angle to which it would be milled) . THE CALCULATION OF SOLID ANGLES 147 Fig. 101 Fig. 102 3. Find the true angle a between the sloping face and the base of the block shown in Fig. 102. 4. A square pyramid is 40mm high and has a base 30mm square. Calculate (a) the angle at the apex between two opposite faces, (b) the angle between two adjoining faces as mea- sured perpendicular to the sloping edge bounding them. 5. A piece of sheet steel 240mm x 120mm is bent as in Fig. 103. Determine the base, height and vertical angle of a triangular piece of material which will fit into the angle as shown dotted. Fig. 103 Fig. 104 6. In Fig. 104 calculate (a) the angle between the corner AB and the top face of the block, and (b) the angle between the two sloping faces as measured perpendicular to AB. Mechanical principles - I Vectorial representation We are familiar with the numerical representation of quantities and with the processes to which we have to submit our calculations to obtain the desired result. In the study of some parts of mechanics it is an advantage to be able to express certain quantities in the form of vectors. When anything has amount and direction it can be represented by a vector, and a vector is a line, the length of which represents the amount of the quantity being represented, and whose direction indicates which way the quantity is acting. Thus in Fig. 105, ab represent a vector 3j units long directed in an Fig. 105 upward direction at 45° to the horizontal; cd represents a vector 5 units long directed horizontally from left to right. Addition and subtraction of vectors When we add or subtract numerical quantities we merely add or subtract the numerical amounts, the result being a numerical sum or difference, as the case may be. When adding or subtracting vector quantities, however, we have to take into account not only their numerical value but also their direction. This is achieved if we observe the following rules: Adding To add two or more vectors, draw the first in the direction of its arrow, continue the second one on the end of the first, the third on the end of the ADDITION AND SUBTRACTION OF VECTORS 149 second, and so on. The sum of the vectors is the vector joining the begin- ning of the first to the end of the last in the series. The arrow on the sum vector must be in the same general direction as those on the vectors being added. This will be understood from the vectors shown added in Fig. 106. Draw ab equal and parallel to vector A ; on the end b drawee equal and parallel to B and from c draw cd equal and parallel to C. The sum of A, B and C is the vector ad, and its arrow is as shown. Fig. 106 The chief mistake made in adding vectors is to add the succeeding vector to the beginning instead of to the end of the previous one. This can be avoided if the reader observes the following rule: When adding vectors, do not remove the pencil from the paper until the end of the last one is reached. Subtracting We know that A - Bis the same as A + ( - B) . Hence to subtract a vector B from another one A we may add —BtoA.A minus vector is a positive one with the direction of its arrow reversed. This if a > b is a vector representing + A, then b < c will represent — A. The following example will illustrate the subtraction of vectors: Example 1 . Subtract a vector of 5 units horizontal L to R, from a vector of 4 units directed at 45° to the NE, and then add a vector of 6 units vertical downwards. The vectors are shown at Fig. 107 (a), and the problem is B -A + C 150 MECHANICAL PRINCIPLES — I A=5 +c ,r C=6'< Fig. 107 It does not matter in which order we deal with the vectors, providing we observe the rules for their addition or subtraction. In this case we have to reverse the arrow of A and then add them all together. This is shown at (b) and the result is the vector ad. To convince the reader that the result is unaffected by the order in which the vectors are drawn, the diagram has been re-drawn at (c), where the vectors have been taken it the order C B A instead of B A C, as at (ft). The result is the same for each case, since eh and ad are equal in length and direction. Exercises 6a [Where an angle is specified, it refers to the angle made with the horizontal or vertical line drawn to the beginning end of the vector.] 1 . Draw vectors to represent the following: (a) 7 units, vertically upwards; (b) 8-5 units horizontal L to R; (c) 3 units upwards at 30° to L of vertical; (d) 64 units downwards 15° to R of vertical; (e) 7-2 units R to L at 30° above horizontal; (/) 5 units L to R at 45° below horizontal. 2. Add together the following vectors: (a) 6 units vertically upwards, to 4 units horizontal L to R. (/?) 8-2 units horizontal R to L, to 7-1 units vertically downwards. (<•) 4-5 units downwards, 10° to R of vertical, to 6-5 units vertically downwards. (if) 7-2 units R to L 20° above horiz, to 6-5 units L to R, 45° below horiz. (<?) 6-3 units vert upwards to 6 units horiz L to R to 6 units R to L, 30° below horiz. APPLICATIONS OF VECTORS 151 (/) 5 units horiz L to R, to 5 units vert, downwards, to 8 units L to R, 30° above horiz. 3. In Ex. No 2, subtract the second vector from the first in (a), (b), (c) and (d). 4. In Ex. No 2, (e) and if), subtract the third vector specified from the sum of the first two. 5. A horizontal vector, ab, 8-6 units long, arrow L to R, represents the sum of two vectors, ac and cb, where acb- = 90°. If ac is L to R, 40° above the horizontal, find the values of ac and cb. 6. A vertical vector, ab, 10 units long, arrow downwards, represents the difference of two vectors (ac - cb). Vector ac is 8 units long, downwards, 30° to R of the vertical. Find the value of vector cb . 7. A vector ab, 15 units long, L to R, 45° above the horizontal, represents the sum of two vectors, ac and cb. The angle between ac and cb is 30°, and ac is upwards, 30° to R of the vertical. Find the values of ac and cb. 8. When two vectors, ab and be, are added, the result is a horizontal vector 10 units long, L to R. When be is subtracted from ab the result is an upward vector, 30° to the R of the vertical, 10 units long. If the angle between ab and be is 90°, find their values. Applications of vectors We will now consider some of the examples which occur in practice, requiring the use of vectors for their solution. Forces In order to specify a force completely we must know its amount, its line of action and its direction. Force may be represented vectorially since the length of the vector may represent the amount of the force, the inclination of the vector may represent its line of action and the arrow head will show the direction. When problems arise where a number of non-parallel forces are acting at a point, the solution can be arrived at vectorially, since this method takes into account the angular effect as well as the magnitude of the forces. A simple example will probably make this clear. Suppose that a pin is being driven into a hole by a force F, applied on an angle as shown in Fig. 108 (a). We know from experience that if F were large enough it would eventually drive the pin home, although it would not do so as well as if it were acting vertically. We also know that if F were sloping too far over towards the horizontal, it would bend the pin . We may say, therefore, that F is equivalent to the combination of a vertical force and a horizontal one, and Fig. 108(6) shows the forces acting on the pin: Q is the pressure of the side of the hole and R is the resistance tending to prevent the pin from entering the hole. By drawing a vector diagram we may find Q and F 152 MECHANICAL PRINCIPLES — I Fig. 108 if we know the amount and direction of F, because we know that F — Q + R. The diagram is shown in Fig. 108(c). The vector for F is drawn parallel to F and equal to its amount to some scale. Since only 3 forces are acting, the diagram is a triangle, and this is completed by making one side parallel to Q and the other parallel to R, Q, being drawn from one end of F and/? from the other./? represents the driving in effect, and Q and bending effect ofF. The reader will observe that here the arrows follow round the diagram, whereas when we were discussing the addition of vectors they did not. Here we are dealing with vectors representing a set of forces which are in balance amongst themselves, whilst before we were finding a vector which represented the resultant of a number of others. If F were representing the resultant, or the net effect of Q and R, its arrow would point upwards, but as it represents a force which balances the other two, then its direction must be reversed. The only difference between the resultant of a number of forces and the single force that will balance them is that the balancing force is opposite in direction to the resultant. Example 2 . A round bar of metal which exerts a vertically downward force of 100N is resting in a 90° vee block. Find the load on the sides of the block. In all problems of this type, a clear conception of the forces acting should be gained before attempting the vectorial solution. Also, in all APPLICATIONS OF VECTORS 153 100N Fig. 109 vectorial problems, the data or details must be drawn out to scale in order that the vectors may be drawn parallel to the quantities they represent. In this case the weight of the bar (100N) acting vertically downwards is being balanced by the forces R t andR 2 exerted on the bar by the vee block. As the vee block is symmetrical about the centre line, the forces R x andi? 2 will be equal. The bar, of course, exerts equal and opposite forces, R Y and R 2 , on the block (Fig. 109a). In drawing the vector diagram, first draw a vertical vector to represent the weight of the bar acting downwards. From each end of this draw vectors parallel to the balancing forces acting. The lengths of these will represent the magnitude of the forces. This is shown in Fig. 109(b). It will be seen that in the vector triangle abc, since each side of the vee block slopes at 45°, the angles abc are each 45° . Hence be = /?, = ac = R 2 = ab sin 45° = 0-707 ab = 0-707 x 100 = 70-7 N Example 3. A casting whose weight exerts a downward force of 1000N is slung by chains as shown in Fig. 1 10(a). Find (a) the tension in the chains, (b) the least angle a between the chains and the casting if the tension in the chains is not to exceed 1600N. 154 MECHANICAL PRINCIPLES — I 1000 i k b (*) Fig. 110 (a) The forces acting at the ring where the chains meet are (1) lifting chain pulling upwards, (2) sling chains pulling in their directions. These forces are shown by the arrows. The diagram must be set out to scale so that vectors may be drawn parallel to their forces; this has been done for Fig. 1 10(a) to save an addi- tional diagram. We may now draw the vector diagram, and this is shown in Fig. 1 10(b). ab is the vertical vector for the upward pull of the lifting chain, whilst ac and be are drawn parallel to the sling chains. Upon measurement, ac and be are found to have a length representing 700N, which is the force in the chains. (b) If the tension in the chains is 1600N, the vector diagram will be as shown in Fig. 1 10(c), and by measurement the angle a is found to be 19° . The reader will observe that the smaller the value of a, the greater will be the load on the sling chains. Exercises 6b 1. The leadscrew of a lathe is threaded 5 mm lead, and is connected to the spindle by a f gear ratio. The feed shaft is set to give a feed of 1 mm per rev to the cross-slide. If the nut and the cross-slide feed (outwards) are engaged at the same time, determine the actual move- ment of the tool point. CONDITIONS FOR EQUILIBRIUM 155 2. A fitter holds a chisel at an angle of 40° to the horizontal, and strikes it a 50N blow with a hammer. Find the force tending to drive the chisel horizontally, and that tending to drive it vertically into the metal. 3. A bar of steel, which exerts a vertically downward force of 300N, rests symmetrically on a pair Of 90° vee-blocks. (a) Determine the reaction between the bar and the block at an area of contact (b) If contact takes place on an area of size 50 mm x 0-08 mm, what is the contact pressure in N/mm 2 ? 4. A wheel, 60mm diameter, rolls along a fiat surface at 0-06 m/s. Determine the actual speed and direction of a point on its circumference, and level with its centre. 5. A bar of steel 2m long and 31 kgf (304 N) weight, is lifted by a chain attached to its ends. The total length of the vee formed by the chain is 2-5 m. Determine the tension in the chain. 6. A casting weighing 20 kgf (196 N) is suspended by a chain and is being pulled to one side by another chain attached to it. If, when the first chain is inclined at 30° to the vertical, the angle between the chains is 105°, find the tension in the chains. 7. A casting of weight equivalent to 2000N is raised by driving 4 wedges under it. If the angle of each wedge is 25° and the weight is equally distributed between them, find the force tending to push the wedge out, and the pressure perpendicular to the wedge surface. [Neglect friction between the wedge and casting.] Conditions for equilibrium For a body to be in equilibrium under the action of 3 forces, the forces acting upon it must satisfy one of the following conditions: (a) They must be parallel, or (b) If not parallel they must meet at a point, and their vectors, when drawn, must form a closed figure. As we shall be dealing with parallel forces later, we will consider case (b). It is not always obvious, upon the examination of a problem, in which direction all the forces are acting. Generally the points which they are being applied can be picked out easily, as can the directions of at least two of them. When we know that to be in equilibrium, non-parallel forces must meet at a point, we can generally find this point by using what information we have, and employ it to help in the solution of the problem. In this connexion we might remind the reader of the following: (a) The weight of a body always acts vertically downwards through its Centre of Gravity, and if the mass is m kilogrammes, the weight can be taken as 9-81 m newtons. (b) When friction is neglected the pressure between two bodies acts at the point of contact and in a direction perpendicular to a tangent drawn to the surfaces in contact. 156 MECHANICAL PRINCIPLES — I Example 4. A bar of metal, mass 15 kg (147N weight), is rested on its end and leaned against a wall so that it is inclined at an angle of 60° to the floor. If the friction between the end of the bar and the wall is neglected, find the force on the floor and against the wall. The forces acting on the bar are (1) The gravitational pull of its weight, acting downwards through its centre; (2) the reaction of the wall acting horizontal (since friction is neglected); (3) the reaction of the floor. Of these we know the amount, direction and line of action of ( 1 ), the direction and line of action of (2), and the point of application of (3). If lines representing ( 1 ) and (2) are drawn on the diagram they meet at O [Fig. 1 1 1 (a)] . Obviously, if the forces acting must meet at a point, the third force must pass through O, and since it must also act at the end B of the bar, its line of action is OB. Having thus determined the directions of all the forces acting, and knowing that they must form a closed vectorial figure, we may now draw the vector diagram. I 147 NT Fig. Ill Draw ab to represent the weight of the bar, be parallel to OB, and ca parallel to AD. From the diagram we find that reaction of floor (be) = 153N and load on wall (ca) = 40N. Example 5. A bell crank lever ABC is pivoted at B and the forces acting at A and C are as shown in Fig. 1 12(a). Find the force on the pivot B. FORCES ACTING ON A CUTTING TOOL 157 If the lines of action of the forces at A and C are continued they meet at D [Fig. 112(6)], so that the third force acting on the lever must pass through D. Since the lever is supported at B, this force must also pass through B, so that the forces acting on the lever are as shown by the arrows. >A 200N 60 («) B 120 C 100Nrr Fig. 112 The vector diagram of forces is shown at Fig. 1 12(c) and from it we see that force acting at B = ac = V200 2 + = 224N 100 2 Forces acting on a cutting tool We have noticed previously that when a lathe tool is cutting there are three perpendicular forces acting on it. These are shown diagrammatically in Fig. 1 13(a), where Cis the vertical cutting force, Fthe feeding force, and// the horizontal pressure of the work. By means of vector diagrams we may determine the resultant of these forces. Since the forces are acting in two planes we must solve the problem in two stages. Let us find the resultant of F and C first. This is shown by the vector diagram abc at (b), and R FC , represented by ac, is the resultant, act- ing in the vertical plane containing Fand C. We may now imagine the tool being acted upon by the forces R FC and //, acting in a plane inclined at to 158 MECHANICAL PRINCIPLES — I Fig. 113 the vertical as shown at (a). Combining these in a second vector diagram, we obtain the final resultant force on the tool (R) as shown at (c).R lies in a plane sloping at 6 to the vertical and its line of action is inclined at a to the line of H . A diagrammatic sketch of the forces and their resultants is shown at (d). The reader will observe that we might solve this problem by calculation f alone, since from Fig. 113(6) F 2 + C 2 = R 2 , and-^ = tan 6; and from C (c)R 2 = R 2 + H 2 and tan a =^ H In all such cases, however, it is advisable to be able to visualise the conditions and not to calculate a result blindly from a formula. Example 6 . The forces acting on a lathe tool are C = 1000N, F = 100N and H = 400N [Fig. 113(a)] . Find the resultant force on the tool. THE BALANCING OF WORK ON LATHE FACE-PLATES 159 As we have discussed the problem diagrammatically, we will solve this problem by calculation. Referring to Fig. 113(6), (c) and (d). R fc = F 2 + C 2 = 700 2 + 1000 2 R FC = V700 2 + 1000 2 = . . F 700 n _ tan0 = C = TOOO = a7 from which 6 = 35°. 1220N R* = R FC * + H 2 1220 2 + 400 2 R = V1220 2 + 400 2 = 1285N tan a = R. 1220 = 3-05 H 400 from which a = 71° 51' Hence the resultant is a force of 1285N, and the angles a and 6 in Fig. 113(fi0 are 71° 51' and 35° repectively. The balancing of work on lathe face-plates When a mass is rotating at a certain distance from the centre of rotation the disturbing effect due to its being out of balance is proportional to the mass and to its distance from the centre of rotation. Furthermore, the disturbing effect is always directed from the centre to the masses. Thus if Fig. 1 14 represents a rotating plate carrying rotating masses m, and m 2 at radii r l and r 2 , the out-of-balance forces are proportional to m l r 1 andm 2 r 2 and are directed from the axis (O), through the mass centres. We are thus able to deal with such problems vectorially, provided we draw our vectors equal in length to the product mr. Fig. 1 14 160 MECHANICAL PRINCIPLES — I Example 7. A rotating plate carries masses of 15 kg and 20 kg placed at radii 80 mm and 100 mm respectively. The angle between the masses is 120°. Find where a 22 kg mass must be placed to balance the system. The two masses are shown at Fig. 115(a). Wrfor pl}~** _ . <t of Balance Weight Fig. 115 The products of their mass and radii are 15 X 80 = 1200 20 x 100 = 2000 The vector diagram is now drawn with the vector lengths proportional to 1200 and 2000, the directions of the vectors being parallel to radii join- ing the c entre O to each respective weight . This is shown at abc in Fig .115 (b) and ca is the vector representing the product mr for the balancing mass. From the diagram the length of ca is 1700 units, and since for the balance 1700 weight m = 22 kg, r 22 = 77 mm. The position of the balance weight relative to the other two is found by drawing a line through O parallel to ca . This is shown dotted on Fig . 1 1 5(a) . Hence to balance the masses given, the 22 kg mass must be fixed at 77 mm radius, and 97° from the 15kg mass. When working out the balancing of irregularly shaped castings, the point or points must be determined where the mass of the casting is acting (the Centre of Gravity). If the metal is all concentrated together, the centre of gravity may be determined as a single point, but if the casting consists of lumps of metal concentrated at different positions, a better method would be to estimate them as separate units of mass. APPLICATIONS OF VECTORS 161 Exercises 6c 1. The forces acting on a lathe tool are 1200N vertical, 800N along the axis of the work and 500N outwards, perpendicular to the axis of the work. Find the amount and direction of the resultant force. 2. If the tool in question 1 is 20mm deep, how far back from its point does the line of action of the resultant force intersect its base? 3. The helix angle of the tooth of a spiral milling cutter is 20°. When the tangential cutting force perpendicular to the cutter axis is 850N, find the end thrust and the force acting perpendicular to the tooth face. 4. The arms of a bell crank lever are 60mm and 40mm long. When the long arm is inclined at 45° to the upwards vertical, the short arm is below the centre line, and with the lever in this position a vertical upwards force of 120N at the end of the long arm is balanced by a horizontal force of 180N at the end of the short arm. Find the force acting on the lever pivot. 5. A bar of steel weighing 31 kgf (304 N) rests with one end in the corner between a wall and the floor and is inclined at 30° to the horizontal. It is held in this position by a rope attached to the outer end, the rope making an angle of 90° with the bar. Find the tension in the rope and the force where the bar rests in the corner. 6. A countershaft is driven by a horizontal belt and the down belt makes an angle of 75° with the top one. When the tension in each side of the driving belt is 200N and in each side of the down belt 120N find the resultant force on the shaft. 350 N rmXJ r- y \Z50N Fig. 116 7. Fig. 1 16 is a diagrammatic sketch of a gear drive, and forces of 350 N act on the teeth of the 100mm gear as shown. Find the resultant thrust on the bearings of this gear. 8. A toggle press mechanism is as shown in Fig. 121. When the horizontal force at C is 250N and the angle CA 2 D is 5°, find the vertical force at the ram. 9. A casting is bolted to a lathe face-plate, the total mass being equivalent to 162 MECHANICAL PRINCIPLES — I masses of 50kg at 200mm radius and 60kg at 260mm radius. The angle between these is 120°. Find the radius and position of a 50kg mass which will give balance. 10. For the mechanism shown at Fig. 126, if a load of lOkN acts vertically down- wards at E, what is the thrust in link BC? 11. A large CI pulley has two out-of-balance bosses on it each 60 mm diameter and 40mm thick. Their radii are A = 260mm and B = 360mm, and the angle between radii drawn to A and B is 110°. Find the volume and position of a lead balance weight to be attached to the pulley at 360mm radius. [Density CI = 7-5g/cm 3 , lead = ll-3g/cm 3 ], Vector diagrams of velocity Velocity, as well as force, may be represented by a vector since it has amount, line of action and direction. The application of vector velocity diagrams is very useful when studying the speeds of points in machine mechanisms, and for that reason it is worthy of consideration. Before going further into the subject it will be well to impress certain fundamental points on the reader's mind, as the success or otherwise of his further study will depend upon his appreciation of them. Fig. 117 (1) When an object is rotating in a circle its velocity at any instant is directed perpendicular to the radius upon which it lies. If co rad/s is the speed and r is the radius, then the velocity is cor. For most engineering applications the rotational speed is usually quoted in rev/min and the radius in millimetres. For general purposes the most convenient unit for velocity is metres per second, and hence if N = speed in rev/min and r = radius in millimetres ,, 2nrN 7idN A . , then V = 60000 = 60000 metres / second VECTOR DIAGRAMS OF VELOCITY 163 (2) Whatever may be the motion of a rigid rod, the only velocity that one end may have relative to the other is perpendicular to the rod. (The "relative" motion of one body to another is the motion the first would appear to have to an observer situated at the second one.) Fig. 118 In Fig. 1 18, AB is a rod having any motion whatsoever. To an observer at A, B can only appear to move perpendicular to AB. If it could move in any direction other than this, it must either approach nearer to A or recede from it. Both of these are impossible since AB is rigid and of fixed length. (3) When a part of a machine is guided in slides it can only move parallel to the slides. We will discuss vector velocity diagrams by working one or two pro- blems, and the reader is advised to take particular note of the system of lettering adopted. Example 8. ABC is a slider crank mechanism. Find the speed of C for the position and values given (Fig. 119). Speed of B = 27irN 2 x 22 x 60 x 500 60000 = 3-142m/s 7 x 60000 500 , rev/min Fig. 1 19 164 MECHANICAL PRINCIPLES — I From the diagram we know the following: (1) B is moving perpendicular to AB at 3-142m/s per sec, (2) Relative to B, C can only move perpendicular to BC, (3) Relative to A, C can only move horizontal. 306 c Fig. 120 The velocity diagram is shown at Fig. 120 and the explanation of its construction is as follows: (Arrows are shown on the vectors, but in practice these are omitted.) ab represents the velocity of B relative to A (i.e. the velocity B would appear to have to an observer at A). The relative velocity of C to B is perpendicular to BC, hence from b a line is drawn perpendicular to BC, and c must lie somewhere on that line. But relative to A, C can only move horizontal, hence c must lie somewhere on a horizontal line through a. The point c is therefore given by the intersection of the two lines drawn, and abc is the velocity diagram. For the given position C is moving towards A at a speed of 3-06 m/s, scaled off from ac. [Note in the lettering of velocity diagrams that ab = velocity of B relative to A and not of A relative to B as might have been expected. Similarly for ac, be, etc.] Example 9. The mechanism of a toggle press is shown in Fig. 121 (a). Find the speed of the ram for the position shown. The figure is drawn to scale. When the frame (the fixed element) appears in more than one place it should be given the same letter with small figures. In this case A, and A 2 are Iboth fixed frame points and are lettered accordingly. The vector velocity diagram is shown at Fig. 121 (b). .. . . + e D InrN 2 X 22 X 60 X 50 Velocity of B = ^^ = 7x60000 = 0.314m/s VECTOR DIAGRAMS OF VELOCITY 165 -260 Fig. 121 Draw ab perpendicular to A,B to represent the velocity of B relative to A, . From b draw a line perpendicular to BC, and from a draw per- pendicular to A 2 C. The intersection of these lines gives the point c. From c draw a line perpendicular to CD and from a draw a vertical line (velocity of D relative to A is fixed as vertically) . The intersection of these gives point d. The vector ad represents the relative velocity of D to A (the frame). This is shown to be downwards and scales off from the diagram to be 0-052 m/s. If the speed and direction of a point on any lever is required, it can be found immediately from the vector diagram, since lines on the vector diagram are images of corresponding ones on the mechanism. Thus for a point E, § the distance from C to D, locate the point e, § from c to d, and ae represents the velocity of E. There are one or two additional points to be explained in connection with velocity diagrams, and these will be covered by the next example. Example 10. The mechanism for a slotted-link shaping machine quick- return motion is shown drawn to scale in Fig. 122 (a). Determine the speed of the ram for the position shown. The crank A,B rotates, and the block B moves up and down in the slotted lever A 2 D . This lever is thus made to oscillate about its pivot A 2 2, and through the link DE drives the ram of the machine. 166 MECHANICAL P. When dealing with a mechanism in which the end of one lever slides in Or on another lever which itself moves, the lettering up should be arranged as shown. B represents the block, and C represents a point on the lever A 2 D. B is rotating about A, whilst C is rotating about A 2 . The reason for doing this will be seen when we discuss the velocity diagram. vi •♦ f D 2 x 22 x 80 x 40 .,__ . Velocity of B = 7 x 600 oo = °^ 5 ^ Ram 'in andBJock) C( Point on Lever \ /ajd; Fig. 122 In Fig. 122 (b) ab represents the velocity of B relative to A. Now the velocity of C relative to B (or B to C) is along the lever A 2 D. From b draw a line parallel to A 2 D and from a draw a line perpendicular to A 2 D (since relative to A, C is moving perpendicular to A 2 D). The inter- section of these gives the point c. The line ac now represents the relative velocity of C to A. The velocity of D relative to A will be a continuation ac oiac to d such that — j ad A 2 C A,D because C and D are on the same lever and their movements are proportional to their respective radii. (For this FORCES ACTING IN A MECHANISM 167 purpose C is assumed to be coincident with B, as it is only an imaginary point introduced for this purpose of the construction.) When d is thus located, draw a line through it perpendicular to DE, to represent the velocity of E relative to D and draw a horizontal line through a to represent the relative velocity of E to A. The intersection of these gives the point e. From the lettering it will be seen that E is moving to the right and the scaled length of ae gives its speed to be 0-353 m/s. The speed at which the block B is sliding in lever A 2 D is given by the length be on the velocity diagram. Forces acting in a mechanism The vector velocity diagram for a mechanism may be used to estimate the force acting at a certain point when the force at another point is known. Let us assume that for a mechanism we have the velocity of two points A and E, and the force acting at A. If v A and v E are the velocities, and F A and F E the forces, respectively, then assuming A to be the energy input point and E the point of output, the input rate of work at A will be F A v A and if the machine is 100% efficient, this will equal the output rate F E v E . We can assume an efficiency r] and then we shall have _ Output _ F E v E v ~ Input ~ F A v A or, since we know F A , v A and v E , F _ t ?( F a v a) E V E Example 11. If the torque input to the toggle press in Example 9 is lOONm and the efficiency of the mechanism is 60%, estimate the pressure on the ram for the position shown. Since torque = (Force)(radius), and r = 60 mm = 0-06 m 100 = (Force)(0-06) and the force at the end 100 of the crank = j^-^ = 1670N U-Uo Here we have that v B = 0-3 14 m/s, v D = 0-052 m/s, and F B = 1 670 N Hence efficiencv - FpVp 0-6 = F d x °-° 52 Hence emciency - — . uo im x Q314 From which F D = ' 6 X ^ ' 314 = 6000 N 168 MECHANICAL PRINCIPLES — I Exercises 6d 1. In a slider-crank engine mechanism similar to Fig. 119 the crank AB is 60mm long and the connecting-rod BC is 220 mm. Find the speed of C when AB is rotating at lOOOrev/min and (a) the angle BAC is 45°, (b) angle BAC = 120°. 2. For the problem in Question 1, find the velocity of the mid-point of BC, when angle ABC = 90°. 3. Fig 123 shows a quick return shaping machine drive in which cutting takes place when C moves to the left. Find (a) The ratio cuttl "g timc t b \ The speed of B in rev / min return time if C is to have a velocity of 0-24 m/s when at the mid-point of its cutting stroke. Fig. 123 4. In Fig. 124 crank AB revolves at 200rev/min and CD is caused to rock backwards and forwards. When AB is at 30° to the horizontal as shown, find the speed of C and of E, the mid-point of BC. 5. In Fig. 125 A and B are two blocks which slide in slots at right angles. If AB = 240mm and AC = 80 mm, find the speed and direction of C when angle OBA = 40° and A is moving downwards at 0-24m/s. Fig. 125 Fig. 126 APPLICATIONS OF VECTORS 169 6. In Fig. 126 crank AB rotates as shown and CDE is a solid bell crank lever. Find the speed of E when AB is at 20° to the vertical. 7. In a toggle press mechanism similar to Fig. 121, A,B = 40mm, BC = 240mm, A 2 C = CD = 160mm. The vertical centre lines are 280mm apart and the centre line of A, is 120mm below the centre line of A 2 . When B is rotating at 60rev/min and angle CA 2 D is 10°, find the speed of D. 8. In the last problem if the torque on A,B is 100 Nm, find the load at D if the overall efficiency is 60%. 9. In the press mechanism shown in Fig. 127, find the speed of E when D is moving upwards at 0.02m/s (A,E is horizontal). If in this position the load at D is 4000N, what force can E exert if the overall efficiency is 50%? [CA,E is a solid lever.] Path of C A,B = 170 ~i^ \ A 2 C = 120 ^> EC = 300 Fig. 127 Fig. 128 10. In the Whitworth quick-return mechanism sketched in Fig. 128, B rotates about A, and slides along the lever CD. CD is pivoted at A 2 . Time of outer stroke, ^ , „w. jn-u ad* Find (a) the ratio for E, and (b) the speed of E when A,B is at Time of inner stroke 45° as shown. 11. In Fig. 129 AB is a door hinged at A. CD is a spring-loaded arm for closing the door and hinged at C. If B is moving at 0-36 m/s, find the speed at which D is sliding along the door when the door has opened 45° . Fig. 129 170 MECHANICAL PRINCIPLES — I The moment or turning effect of a force If a force acts on a body, and the effect of the force is considered relative to some point not on its line of action, the tendency is for the force to rotate the body about the point. This tendency is called the moment of the force about the point. Thus in Fig. 130, if the effect of the force F is considered relative to the point 0, F tends to rotate the body about O . The numerical value of the moment of F about O is found by multiply- ing F by the perpendicular distance from O to its line of action. Thus the moment of F about O = Fx. If F were rotated to act in some other direc- tion (e.g. as shown dotted), then its moment would h^F, (perpendicular distance) = F,_y. To calculate turning moment we multiply a force by a distance so that the unit of turning moment is usually newton metres (Nm) (although on occasions a unit such as Nmm may be more convenient to save converting and re-converting units of length). Fig. 130 Fig. 131 When two equal and opposite forces act on either side of a pivot they constitute what is called a Couple, and the turning moment exerted by a couple is given by one of the forces multiplied by the perpendicular distance between their lines of action (Fig. 131). Total moment of forces P = Fr + Fr = 2Fr = F(2r) = F(perp distance between forces). A good example of a couple is given by the forces applied to a tap wrench or die stocks when cutting a thread. The couple applied is equal to the turning moment resistance at the cutting edges of the tap or die. CONDITIONS FOR EQUILIBRIUM WITH MOMENTS ACTING 171 Moment of a force about a point on its line of action Since moment = (Force)(distance), if the distance is zero, then the moment will be zero if the point lies on the line of action of the force. Thus in Fig. 130 the moment of F about A = F x O = 0. Conditions for equilibrium with moments acting The reader will have observed that a moment may be directed clockwise or contra- clockwise. The condition for a body to be in equilibrium isthat the sum of the clock- wise moments about any point must be equal to the sum of the contra-clockwise moments. When this condition is satisfied there will be no unbalanced turning effort causing the body to rotate. Example 12. A shaft is 7-5 m long between the bearings and carries 3 pulleys spaced at 2 m, 3 m and 6 m from the LH bearing. The downward force on the pulleys due to the belt drives is 300 N, 250 N and 350 N respectively. Calculate the load on each bearing (Fig. 132). -7-5m Fig. 132 Since the loading on the shaft by the pulleys is downwards the forces exerted on the shaft by the bearings will be upwards. Call these R A andi? B . We thus have the shaft acted upon by a system of parallel forces, and since it is in equilibrium, the conditions to be satisfied are: (1) Total upward forces (2) Clockwise moments Total downward forces. Contra-clockwise moments. If we take moments about the bearing A, we can neglect the force acting there since it will have no moment about that point. 172 MECHANICAL PRINCIPLES — I Hence by moments about A Clockwise moments = 300 x 2 + 250 x 3 + 350 x 6 = 600 + 750 + 2100 = 3450Nm Contra-clockwise = R B X 7-5Nm These must be equal 3450 7-5 R B = 3450 and R B = 460 N Also, since upward forces = downward forces: 300 + 250 + 350 = 900N *a + *b R A = 900 R, 900 - 460 = 440 N Example 13. A system of levers is shown in Fig. 133. Calculate what load hung at W may be balanced by a force of ION at A. [Neglect friction and the weight of the levers.] R H 20 h T \ ^ir? \ 240 200 J$2- Wu kip \15 a ^ "10N Fig. 133 First let us consider the top lever, which will be balanced by the force F in the connecting link and the ION at A. Taking moments about B, and working in units of N and mm, we have Clockwise = 10 x 240 = 2400 and contra- clockwise = 20F Hence 2400 = 20 F and F = 120N This will pull down on the top, and upwards on the bottom lever. Now consider the bottom lever and take moments about C. Clockwise = F x 215 Contra-clockwise = W x 15 EXAMPLES INVOLVING MOMENTS 173 These must be equal, and since F = 120 120 x 215 = \5W W = 12 ° * 215 = 1720N The actual value of W would be affected by friction, the weight of the levers and of the stirrup. If the lever weight is appreciable it can be allowed for by assuming the weight of each lever to be acting as a down- ward force approximately at the lever centre. For example, if each lever weighs ION and the weight is assumed at the lever centre, we have for the top lever: 20 x F = 10 x 240 + 10 x 110 = 2400 + 1100 = 2500 F = 175N For the bottom lever F x 215 = 15W - 10 x 107-5 and since F = 175N 175 x 215 = 15 W- 1075, 15 W = 37 625 + 1075 = 38 700 y-Jg™-2580N To obtain an exact calculation, the centre of gravity of the levers could be found by balancing on a knife-edge. The weight is then taken as acting through the centre of gravity. Example 14. If in Example 6, p. 158, the work is 240mm long between centres, and the cutting tool is 60 mm from the tailstock centre, calculate the forces acting on the centres. The forces acting on the work are as follows: (1) Vertical upwards force of 1000N. (2) Horizontal force of 400 N directed away from the tool. (3) Longitudinal dorce of 700 N towards the headstock. These are shown in Fig. 134 (a). Let us consider the vertical force first. This will cause an upward force on each centre and an equal and opposite downward force on the work by the centre. Taking moments about centre A, and working on units of N and mm we have Clockwise moment = (vert force at B)240 174 MECHANICAL PRINCIPLES — I 1000N Headstock (a). Fig. 134 (b). Contra-clockwise moment =1000 x 180 = 180 000 Hence 240 (vert force at B) = 180 000 180 000 Vert force at B = Vert force at A = 240 1000 = 750 N 750 = 250N In the same way we have that the 400N force is balanced by horizontal forces of 300 N at B and 100N at A. The 700 N longitudinal thrust is transmitted through the bar entirely to A. We thus have forces acting on the two centres as shown by Fig. 134(b), and as an additional exercise the reader should combine these vectorially to find the resultants. Exercises 6e 1. A bell crank lever has equal arms of 120 mm at 90°. When the lever is pivoted with one arm vertically upwards, a weight of 120N is hung on the end of the horizontal arm. What force, inclined at 45°, must be applied to the end of the upper arm to balance the lever? 2. A 20 mm hand reamer is being operated by hand pressure at each end of a lever 480mm long. The hole is 20mm long and each of the 6 teeth is taking a cut of 0-04 mm If the cutting pressure at the teeth is 80N per mm 2 of cut, estimate what force must be applied at each end of the 480mm wrench. 3. A bar is 240mm long between lathe centres. When the tool is 90mm from the tail- stock the vertical cutting pressure is 850N. Find the vertical force on each centre due to this. 4. A piece of material is held in a lathe chuck and the point at which cutting takes place overhangs from the centre of the front bearing by 125 mm. If the vertical pressure due to the cut is 1200N, and the bearings are 360 mm apart, estimate the force on each EXAMPLES INVOLVING MOMENTS 175 bearing. [Assume point contact at each bearing and neglect weight of spindle and chuck.] 5. The bearings of a countershaft are 900mm apart. The horizontal force due to the top belt is 450N at 240 mm from the LH hanger and the vertical force due to the down belt is 400N at 720mm from the LH hanger. Find the load on each bearing. 6. A clamp is 120mm long between the centres of the clamping and supporting points. If the bolt is 45 mm from the clamping point, to what tension must it be tightened up in order to apply a clamping load of 1000 N? What will be the reaction at the support? 7. A casting is being raised by a 720mm crowbar which is supported at 40mm from the end where it takes the weight. If the casting has a mass of 612 kg (6000 N weight), and half this is taken by the crowbar, what force must be applied at the end of the bar in order to raise the casting? Blade , P " ot l F/xech 40 Pivot Work'to he sheared Fig. 135 8. A bench shearing machine is shown diagrammatically at Fig. 135. If the shearing strength of steel is 400N/mm 2 , what force F must be applied to shear a piece of material 10mm x 2-5 mm? 7 Mechanical principles — II Friction Friction plays such an important part in a workshop that the calculations connected with it are worthy of some consideration. If we press two surfaces together and attempt to slide one over the other, a resistance is encountered, and this is caused by the friction between the two surfaces. Let us consider a block pressed on to a surface with a force W as shown in Fig. 136. Another force is now applied to the block, tending to slide it from left to right. As soon as this force is applied, frictional resistance comes into action and prevents the body from moving. This resistance is denoted by /and indicated by the arrows at the surfaces. Depending upon W, and on the nature of the surfaces, however, there is a limiting value beyond which / cannot increase, so that if we gradually increase the force tending to slide the block, a point will be reached at which the block will just be on the point of sliding. Let the force then be F, and since the block is just about to slide F = /, the frictional resistance. The block at this point is being acted upon by three forces: (1) W downwards, (2) F horizontal, (3) the reaction of the other surface (say R). The vector diagram is shown drawn in Fig. 136 at abc. F=f Fig. 136 We mentioned earlier, that when friction is neglected the reaction between two surfaces is normal to their common tangent or along ab in Fig. 136. It will now be seen that when friction is allowed for, the reaction moves round so as to oppose motion, and its line makes an angle <j> with the normal line to the surfaces. CLAMPING FRICTION 177 This angle ^ is called the Friction Angle and it will be noticed that . , _ F_ Frictional force ^ ~ W ~ Pressure between surfaces F The ratio -^ is called the Coefficient of Friction and is usually denoted W by fi. *=W F Thus if Tp = tan <j> and also = /u tnen + ± ju = tan <j> . [The reader will appreciate the fact that the reaction R between the surfaces moves round to some angular position, and is directed in such a way as to oppose motion, if he considers where and in which direction it would act in the event of a definite step being raised in front of the block.] Approximate Values for the Coefficient of Friction Nature of surfaces in contact , „ ,T . ^ (Coefficient). Cast iron on cast iron (dry) 0-15 0-20 0-15 0-49 0-56 0-23 0-40 0-33 Ferodo bonded asbestos on steel 0-3- 0-4 [Values of /u depend to some extent upon the pressure between the surfaces and upon the speed of sliding.] Clamping of work Almost all methods of clamping in the shop depend for their hold upon the frictional resistance between the two surfaces being clamped. This applies to work clamped to machine tables, in lathe chucks, work held in vices, and so on. Steel on cast iron (dry) Steel on brass (dry) Cast iron on oak (dry) Steel on leather (dry) Steel on leather (greasy) Oak on oak (dry) Leather on oak (dry) 178 MECHANICAL PRINCIPLES .15* Q 100 40 rffe w: A\ Work T¥ t< 10> w: 7vft W/////////////7/////////////// Fig. 137 Example 1 . A casting is clamped to a shaping machine table by 4 clamps arranged as shown in Fig. 137. If the cutting pressure at the tool causes a thrust of 4500N against the work and if the coefficient of friction between the work and the table is 0-15, what is the least tension to which the bolts be tightened in order that the job may not move under the pressure of the cut. If the sliding force to be resisted is 4500N and /u =0-15, the surfaces 4500N must be pressed together with a force of - 0-15 30 000N Each clamp must therefore exert a pressure of — j — = 7500N In Fig. 137 the forces exerted by the clamp are shown and T = tension in bolt, W A and W B = Pressure exerted on work and packing respectively. We will assume that W A and W B act at the centre of the length being clamped as shown. Then from W A to the centre of the bolt will be 40 - 7-5 = 32-5 mm and from W B 60mm - 5mm = 55mm. Taking moments about B and working in units of N and mm, we have Clockwise moment = T X 55 Contra-clock moment = W A (55 + 32-5) = 87-5 W A . But W A must be 7500N from above, so that putting in this value and equating the moments we have 55T = 87-5 X 7500 T = 87-5 X 7500 55 - 12 000N Example 2. A milling cutter is tightened up between the collars on the arbor and is driven by the friction between itself and the collars. If the FRICTION 1 79 Fig. 138 collars are 20 mm bore and 40 mm outside diameter, calculate the tighten- ing force necessary if the cutter is not to slip when 3-3 kW is being absorbed at 70rev/min [Coeff of friction = 0-15.] If T newton metres is the torque to tbe transmitted, we have Power = T newton metres/ second = Ta> Watts also Power = 3-3 kW = 3300 W, Now N = 70 rev/min cfw 70 x 2tt 22 ,. so that (o = — Z7 r — = -^- rad/s OU J _ Power 3300 x 3 . CAXT T = = « = 450Nm oi 22 This torque must be transmitted by the friction between the collars and the cutter, and slip when it takes place, must occur at two faces: 450 Hence torque transmitted per face = -y- = 225Nm This torque will be developed by a tangential frictional force F which we will assume to act at the mean radius of the collars (Fig. 138). Hence Mean radius = — = — = 15mm = 00 1 5m T =Fr,F = -= _ 2 A1 5 g N = 15 000N r 0-01 5 m Now if W = Force between the collars F . 15 000 A1 _ 7p= p> »- e - w = °' 15 W = l5 ®®° = 100 000N 180 MECHANICAL PRINCIPLES — II Exercises 7a 1. The tailstock of a lathe has a mass of 21.5kg and the coefficient of friction at the slides is 0-122. What horizontal force will be required to slide the tailstock? Determine the amount and direction of the least force necessary to slide it. 2. A disc 240 mm diameter has a ring of ferodo 240 mm outside diameter and 20 mm wide riveted to one face. This is made to press against and drive another steel disc. If the turning moment to be transmitted is 33 Nm estimate the pressure with which the discs have to be pressed together. [Take fi for ferodo on steel as 0-4.] 3. A planing machine is taking a cut on a casting bolted directly to the table. The force of the cut is 7400N and the job is clamped by four clamps. If the coefficient of friction between the casting and the machine table is 0-15, what force must be exerted on the work by each clamp to prevent the work from sliding under the force of the cut? 4. A 300mm diameter brake drum is attached to a shaft the driving pulley of which is 200mm diameter. When two leather-faced brake blocks are pressed against opposite sides of the drum with a force of 200N what force must be applied to the rim of the pulley to turn the shaft. [Take /u for leather on cast iron as 0-45.] 5. The table of a planing machine with weight equivalent to 5000N is supporting a casting which exerts a similar force of 2000N. If "the average speed of the table is 0-2 m/s and for half the time a downward cutting force of 800N is acting, calculate the average power required to overcome friction at the table slides. [Take fi = 0-08.] 6. A block is clampted between the jaws of a milling-machine vice, the force at the jaws being 12 000 N. If the cutter operating on this is 80mm diameter and its speed 52-5 rev/ min, calculate the approximate power being absorbed when the work is caused to slip in the jaws of the vice by the force of the cut. [Take /u at the vice jaws as 0-15.1 Machines and efficiency A mach ine is a contrivance for receiving energy in some form and con- verting it into energy of a type more suitable for the purpose required. Most of the energy available in a machine shop is in the form of rotational energy (line shafting, rotation of driving motors, etc.). A machine, such as a shaper, receives some of this rotational energy at its pulley, and con- verts part of it into the energy contained in the backwards and forwards movement of the ram and part into the various other movements required to transverse the tool across the work. Each element of a machine may be regarded as a little machine in itself. For example, in the mechanism for elevating the knee of a milling machine a torque is applied to the handle, and this torque is converted into the rotation of a nut or screw. This in its turn raises or lowers the table and cross-slide on the vertical slides. An electric motor is just as much a machine as any other, for it takes in electrical energy from the mains and converts it to rotational energy at its driving pulley. MACHINES AND EFFICIENCY 181 For mechanical machines: the Effort is the force applied at the input end of the machine, and the Load is the resistance overcome at the output end. The ratio p~. is called the Mechanical Advantage of the machine and +u~ 4-- Distance moved by Effort . x , . .„,,,, the ratio Distance moved by Load m the same time, is called the Velocity Ratio of the machine. If we called the effort E, and the load W: The work input will be E (Distance moved) and the output in the same time = fF (Distance moved). The Efficiency of the machine will be -= — - — Input _ W (Distance moved) E (Distance moved) W But — = Mechanical Advantage and E Distance moved by W 1 Distance moved by E Velocity Ratio Hence Efficiency = Mechanical Advantage Velocity Ratio If the efficiency were unity, the Mechanical Advantage would be the same as the Velocity Ratio. Actually the efficiency is always less than 1 due to frictional losses, so that the Mechanical Advantage is always less than the Velocity Ratio. Example 3. The knee, cross-slide and table of a milling machine have a total mass of 600 kg, and it is found that a force of 60 N must be applied at the end of the 210mm elevating handle to raise the knee. One turn of the handle raises the knee 2 mm. Calculate the Mechanical Advantage, Velocity Ratio and Efficiency of this mechanism. Effort (E) = 60N Load (WO = mg = 600 x 9-81 = 5886N Mech Advantage = ,. = 98 60 182 MECHANICAL PRINCIPLES — II Distance moved by effort Velocity Ratio = Distance moved by load In X 210 mm 2 mm 660 „„, . Mech.Adv. 98 n*Ao 1^00/ Efficlency - Vel. Ratio = 660 = °'' 48 = M ' 8/ » Example 4. When disengaged from the operating mechanism a grinding machine table can be pushed along by a force of40N. With the mecha- nism engaged, a torque of 0-35Nm at the traversing wheel is required to move it. One turn of the wheel moves the table a distance of 10mm. Find the efficiency of the traversing mechanism. The load on the mechanism = resistance of the table = 40N If 1 turn of the handwheel moves the table 10 mm work done on the load in 1 turn = 40N X 10mm = 40N x 0-010m = 0-4Nm = output Work done by a torque of 0-35 Nm in 1 turn = In X 0-35 Nm = 2-2Nm = input Efficacy -<g*-£ = <M82 -.8.2% The inclined plane We have seen in previous work (Fig. 78a) that a screw thread is an inclined plane wrapped round a cylinder. In order to study the mechanics of the screw, therefore, we must give some attention to the inclined plane. We find that in doing this friction plays an important part and its effects must be allowed for. Tightening up When a nut is being tightened up under a load W, the conditions are equi- valent to pushing a weight W up an inclined plane sloping at the helix angle of the screw. The force F which is pushing W is the tangential force at the mean radius of the screw and is being applied in a horizontal direc- tion. This is shown in Figs. 139 and 140. Now if there were no friction between the block and the plane the SCREW THREAD AS AN INCLINED PLANE 183 Load on Nut(W) Effort (F) Angle of Incline* Helix Angle of Screw Development of Screw and Nut Fig. 139 vector diagram for the forces acting on the block would be as abc; ab = weight of block acting downwards, be = F, the force to pushing it up the plane and ca = the reaction between the plane and the block. We have seen, however, that when friction is present, the reaction R is no longer perpendicular to the surfaces, but is rotated round through the friction angle <j> in the direction to oppose motion. Hence the vector for R will be rotated round to ad, and the force F required to push the block up the plane will become = bd. Lead of Thread j? c Fig. 140 bd /\ Now — = tan bad = tan (or + <j>) ab W tan(ar + ^) and F = Wtan(a + ^) (1) 184 MECHANICAL PRINCIPLES — II In pushing the block from A to B, F moves the distance AC and W is raised to distance BC. Hence the input = work done by F = F.AC and the output = work done on W — W.BC The efficiency {rj) output fF.BC W 4 I BC r . ^ = -77 tan a since F.AC F tan or input F.AC F L" AC But from above: F = W tan (a + (f) and if we substitute this for F we have Wtana tan rv — w. t] = -p tan a = Wtan(a + 0) tana a = Helix angle of screw, i.e. tana = <j> = Friction Angle tan (a + (j>) Lead (2) Mean circum Tan = /u This is an expression for the efficiency of a screw when tightening up and is in terms of /u and the helix angle of the screw. Loosening When a nut is being unscrewed it is equivalent to the load Amoving down the plane. There are two cases to consider: (1) When is greater than a and the load must be pushed down, and (2) when a is greater than </> and a push up the plane is necessary to stop W from sliding down of its own accord. (1) greater than a (Fig. 141). As motion is now taking place in the opposite direction, the reaction line ad swings round to the other side of the normal ac, and the vector d ]? b c Fig. 141 FRICTION AT VEE THREADS bF d Fig. 142 diagram is abd. In this case since <j> is greater than a, ad swings lfound to a position to the left oiab. Force required at mean radius to unscrew nut = F = db — ab tan had = W tan (</> - a) (3) (2) a greater than ^ (Fig. 142). In this case a force = bd will be required to hold the nut fj-om un- screwing. F = bd = ab tan(a — f) = Wtan(a - </>) 185 (4) Vee threads The above reasoning for screw threads assumes that W is the normal load on the thread surfaces. Whilst this is true for a square thread [Fig. 143 (a)] , when a vertical load W is applied to a vee thread the force is resolved into components as shown by the vector diagram abc shown at (b). (b) Fig. 143 186 MECHANICAL PRINCIPLES — II The normal load on the thread surface is now the length ac, whilst be is the force tending to burst the nut. W Hence normal load on thread face — ac — = W sec 6 . cos 6 As the frictional resistance depends on the normal pressure we may modify our previous formulae to apply it to vee threads by modifying the value of the coefficient of friction (/u). If, instead of taking /u we take for it a value equal to /u sec0, we may use the previous formulae as they stand. For a metric thread, the vee-angle is 60°, hence 6 = 30° and sec# = 1-155. Hence for a metric thread take a modified value for the coefficient of friction equal to 1-155 ,u. Friction at a nut face In addition to the friction at the threads, friction at the nut face must be allowed for. Let W = load on nut Hi = coefficient of friction r = mean radius of nut. Then frictional force at nut face = fxW. Torque necessary to overcome this = (force) (radius) = fiWr Example 5. A screw-operated arbor press has a square-threaded screw 50mm dia 5mm pitch single start. If the coefficient of friction at the threads is 0- 10, what load may be applied by the press when an effort of 200N is applied at the end of a handle 200 mm long attached to the screw. Since tan ^ = /u; tan <j> = 0-1 and the friction angle (0) = 5°43' The mean circumference of the thread = ,t(50 - 2-5) = 47-5 n = 149-2 mm If a is the helix angle of the thread Lead 5 nm-»c tana = ~ = . An . = 0-0335 Circum 149-2 from which a = 1°55' The mean radius of the screw = ^ = 23-75 mm so that a force of FRICTION AT A NUT FACE 187 200N at 200mm radius will be equivalent to a force — — — at 23-75 mm radius = 1684N. This is the force F up the inclined plane. Now F = Wtan(« + ^)(Fig. 140) = fTtan(l°55' + 5°43') = W tan7°38' = W X 0-1340 w - orao = 515? = i^ON Example 6. A machine slide weighing 255 kgf (2500 N) is elevated by a 2-start acme thread (29° thread angle) 40mm dia, 4mm pitch. If the coefficient of friction is 0-12, calculate the torque necessary (a) to raise the slide, {b) to lower it. The end of the screw is carried on a thrust collar, 32 mm inside and 56mm outside diameter. Mean dia of thread = 40 mm - 2 mm = 38 mm Mean circum = 38 n = 119-4 mm Lead = 2 x 4 mm = 8 mm g If a = helix angle tan a = -rnrz = 0-0670 a = 3° 50' Since the thread half angle = 14^° we have to use a modified coefficent of friction = 0-1 2 sec 14}° = 0-12 x 1-033 = 0-124 tan^ = 0-124 and ^ = 7° 4' To raise the load we have if F = force at the mean radius F = Wt&n(a + f) = 2500 tan(3° 50' + 7° 4') - 2500 tan 10°54' = 2500 x 0-1926 = 481-5N The torque will be F (radius at which it acts) = F x 38mm = f x 0-019m = 481-5N x 0-019m = 9-15Nm 2 To this torque must be added the friction torque at the thrust collar. Frictional force = 2500^ = 2500 x 012 = 300N Frictional torque = 300 (mean rad of collar) = 300 x ^y5B. = 300N x 0-022 m = 6-6Nm Total torque to raise slide = 9-1 5Nm + 6-6 Nm = 15-75Nm 188 MECHANICAL PRINCIPLES — II To lower the load F = JFtan(0 - a) (Fig. 141) = 2500 tan(7°4' - 3° 50') = 2500tan3°14' = 2500 x 0-0565 = 141-25N Torque = 141-25N x 0-019m = 2-68Nm Adding the friction torque at the collar we have Total Torque to lower the slide = 2-68 Nm + 6-6 Nm = 9-48Nm. Friction of sliding keys It is often necessary in machine-tool construction to slide a collar or wheel along a shaft at the same time as the wheel is being driven by the shaft through one or two sliding keys. The force required to slide the collar or wheel along the shaft is worthy of "consideration as it is affected by the disposition of the keys. In Fig. 144(a) the key is fixed in the shaft which rotates in the direction of the arrow and drives the outer part. Clearances have been exaggerated and it will be seen that the torque is transmitted by a force F at the key and a similar force W acting at the circumference of the shaft. If r is the radius to the centre of the key as shown, and T is the torque being trans- mitted, then T = Fr and F = ~ r The force required to slide the outer member along the shaft will be F/jl + Wfi= IFfx (since F = W). Fig. 144 FRICTION OF SLIDING KEYS 189 In Fig. 144 (b) there are two keys, and if they are well fitted so that each takes an equal share of the load, the torque is given by T = F x r + W x r= F,{2r) [since F l = W,]. T Hence F, = ^-, which is half the force for the case of Fig. 144 (a). Force to slide outer member along the shaft = F y /u + W r /u = 2F 1( u, which is half of that for the case of Fig. 144 (a). Exercises 7b 1. In a hand crane a 20T gear attached to the handle drives an 85T gear on the rope drum. The radius of the drum to the centre of the rope is 35 mm. If the handle is 300mm long, calculate the velocity ratio. If the efficiency is 75%, what force must be applied to the handle to raise a load of 250 kg on the rope? 2. A chain conveyor carries goods up an incline of 28° at a speed of 0-2m/s. If the average mass of the articles carried is 20 kg, and they are spaced at 300 mm centres, calculate the power necessary to drive the loaded conveyor if its efficiency is 75%, the incline carries 60 articles, and 0-4 kW is necessary to overcome friction. 3. The saddle of a lathe is equivalent to a weight of 1600N and 1 turn of the traversing wheel moves it 100mm along the bed. If the efficiency of the traversing gear is 0-7 and the coefficient of friction at the slides 0-10, calculate the force necessary at the rim of a 140mm wheel to move the saddle. 4. A flypress has a screw of 50mm lead, the efficiency of which is 60% . Neglecting the weight of the screw and top arm, what force must be applied at the end of and per- pendicular to the level of 280mm radius to put a force of 2500N on the ram? 5. Calculate the efficiency of a 20 mm square-threaded screw of 5 mm pitch 2 start if the coefficient of friction at the threads is 0-080. What tension may be exerted on this screw by a nut if 120N is applied at the end of a 350 mm spanner? [Neglect friction at the nut face.] 6. The efficiency of a screw and nut is 15%, and the coefficient of friction at the nut face is 0-1. If the lead of the thread is 2mm what force must be applied to the end of a 240 mm wrench to pull up the nut against a tension of 8000 N? [Mean radius of nut = 15mm.] 7. The spindle of a lathe is connected to the leadscrew by the following gears: -=r-: = 30 20 "sn x Z?" ^ l ^ e l ea d screw i s 5 mm pitch, calculate the velocity ratio between the carriage and a point on the rim of an 160 mm chuck screwed on the spindle. If the overall efficiency of the arrangement is 10%, calculate the force necessary at the rim of the chuck to turn the lathe and traverse the carriage against a resistance of 200N. 8. Calculate the efficiency of a M24 thread when the coefficient of friction at the threads is 0-08. [For the M24 thread take the mean diameter as 22 mm and the pitch 3 mm.] 9. Calculate the work done in pushing a slide of mass 100 kg up an inclined plane sloping at 30°, if the coefficient of friction is 0-15, the plane is 3 m long, and the push is applied horizontally. What horizontal effort would be necessary to hold the slide from moving down the plane? 190 MECHANICAL PRINCIPLES 10. A wheel slides along a shaft and is driven by a sliding key. If the shaft is 45mm diameter and the key projects 5 mm from the circumference, calculate the force necessary to slide the wheel along the shaft when 6-6 kW is being transmitted at 315rev/min and H = 0-15. 11. If the shaft in the last example were fitted with two opposite keys projecting the same amount as before, calculate the torque being transmitted when an axial force of 60 N is required to slide the outer member along the shaft, i/n = 01 5.] Bearings The main function of a bearing is to hold and line up the shaft it carries and support the load to which the shaft is subjected. Bearings are generally designed on the basis of the load carried per unit of projected area, and if the length and diameter of a bearing are / and d respectively, the projected area will be / x d (Fig. 145). Fig. 145 If the load on the shaft = W, the intensity of bearing pressure (p) will be Load Area W P= ld The pressure to which bearings may be subjected in practice depends upon various factors, including the speed, method of lubrication, dura- tion of full load operation, materials in contact, and so on. 191 The following table conveys an idea of bearing pressures used: Table of Bearing Pressures Allowable Type of Bearing Pressure N/mm 2 Line-shafting (bronze lined) 0-7- 1-0 High-speed engines: Main bearings 1-0- 2-0 Crank pins 2-0- 4-0 Gas engines: Main bearings 3-5- 5-0 Crank pins 10 -12-5 Punching and shearing machines (low speed intermittent loading) 15 -30 Horizontal turbines 0-3- 0-5 Probably the most severely loaded bearing in the whole of engineering practice is the tailstock centre of a lathe. Let us consider the vertical tool pressure only, and assume the moderate case of a load of 4000 N, with the centre in a hole measuring 5 mm at the large diameter of the countersink. Since the angle of the countersink is 60° the projected area will be i X 5 x 0-866 x 5 = 10-8 mm 2 and the bearing pressure when the tool is cutting close to this centre will be 4000N -r^r r = 370 N/mm 2 , about half the ultimate failing stress of a good 10-8 mm 2 quality mild steel ! The main bearing carries only the same load with an area of probably 3000 to 4000 mm 2 . Example 7. If the bearings for the shaft in Example 12, p. 171 are to be proportioned so that their length shall be 2\ times the diameter, calculate their dimensions if the bearing pressure is not to exceed 0-4 N/mm 2 . The biggest load to be carried = 460 N If L = 2\d, L x d = 2W x d = 2\d 2 = Area of bearing. ^ <• . • Load 460 nm , But area of bearing = =, = 77-r- = llSOmnr b Pressure 0-4 Hence 2\d 2 =1150 <P = il^ = 460mm 2 192 MECHANICAL PRINCIPLES d = \/460 = 21-5 mm L = 2\ x 21-5 = 53-8 mm Bearing friction When a bearing is properly lubricated the two metals forming it are not in contact but are separated by a thin film of oil. The friction now is not that of one metal rubbing on another but is the internal friction of the lubricant itself. It has been found experimentally that the coefficient of friction in a bearing with film lubrication depends upon the rubbing speed and upon the pressure, and the relation between them is P = KV\ where /u = coeff. of friction v = surface speed of shaft in m/s p = bearing pressure in N/mm 2 (i.e. MN/m 2 ) K = a Constant = 0-032 for the usual oils. Work lost in bearing friction The effect of friction in a bearing is to introduce a tangential resistance at the periphery of the shaft. If W = load on the bearing and n = co- efficient of friction, then the tangential resistance will be W = /u. Calling this resistance F, we have F = W/u (Fig. 146). The work lost per second will be Fv and the power lost = Fv watts Fig. 146 STRESS AND STRAIN 193 Example 8. A 50mm dia shaft running at 525rev/min carries a load of 8000N.The bearing is 100 mm long. Estimate (a) the coefficient of friction, (b) the tangential friction resistance, and (c) the power lost in friction. w u *u ♦ 0-032Vv We have that a = P v = J^L = 22 >< 50 >< 525 = 1.375m/s 60 000 7 x 60 000 8000 Bearing pressure (p) = -^ r^r = 1-6 N/mm 2 0-032 x W-375 0-032 x 1-172 _ 0-0234 ^~ 1-6 ~ 1-6 Tangential frictional resistance = 0-0234 x 8000 = 187 N Power lost in friction = 187 N x 1-378 m/s = 257W Stress and strain A material is placed in a state of stress when a load acts upon it, and the numerical value of the stress is given by Load acting F , FT j a a a r-: . p , — r = -r, when F = Load, and A = Area Area subjected to load A The SI unit of stress is therefore one unit of load divided by one unit of area, i.e. the newton per square metre (N/m 2 ). This unit is very small for practical purposes and so stresses will often be quoted in meganewtons per square metre (MN/m 2 ). This unit is quite convenient, since lMN/m 2 = lN/mm 2 and as loads are often quoted in newtons, while dimensions of engineering components are usually quoted in millimetres, if we ever require a stress in MN/m 2 it is useful to evaluate the stress in N/mm 2 , e.g. 45 N/mm 2 = 45 MN/m 2 . 194 MECHANICAL PRINCIPLES — II It is possible that the unit adopted for fluid pressure will be the bar, a. unit inherited from the previous metric system. In some ways this is a more convenient unit since it utilises the cm 2 as the unit of area. 1 bar (b) = da N/cm 2 = 10 N/cm 2 1 hectobar (hb) = 10 3 N/cm 2 As a mental landmark, the bar is very near to atmospheric pressure: 1 bar = 10 N/cm 2 1 atm = 10-13 N/cm 2 The weight of a body is the effect of gravity on its mass. The effect of gravity on a mass of m kilogrammes can be taken as, weight = mass x acceleration due to gravity = mass x 9-81 newtons. As we discussed in Chapter 1 , the reader will find, in his normal life, that the mass of 1 kilogramme is referred to as "weight" or a kilogramme of mass and written 1 kg, with the inference of weight. In our work it is sometimes expedient to express the weight of 1 kg of mass in its kg form and when doing this it should be written kgf . At the same time it should be remembered that 1 kgf =9-81 newtons. When materials are stressed they change their shape: for example, a bar will lengthen under tension or shorten under compression. This change of shape is called strain. We usually express the strain in terms of the natural length ofthe material, so that if abar 100 mm long is subjected to tension and stretches 01 mm, the strain is expressed as j^- = 0-001. Within certain limits the materials with which we have to deal behave in an elastic manner, i.e. the deformation caused by a load vanishes when the load is removed. If, however, the load on a bar is gradually increased, a point is reached beyond which the material will not return to its original shape when the load is removed. This point is called the Elastic Limit ofthe material. For most materials it has been found that within the elastic limit the change in length is proportional to the load producing it: e.g. if 1000 N causes an elongation of 0-05 mm, 2000 will cause 010 mm, and so on. Hence we may say that within the elastic limit: Stretch is proportional to Load STRESS AND STRAIN 195 or, since for the same bar, strain is proportional to stretch and stress to load: Strain is proportional to Stress. Stress This is the same thing as saying: ~- — — = a constant quantity This constant quantity is called Young's Modulus, and has a particular value for every material. It is usually denoted by the letter E. For steel E has a value of about 200 000N/mm 2 . It is difficult at first to visualize and to appreciate what E = 200 000 N/mm 2 signifies. The following way of considering it might help the reader: Stress t= — — means Stress per unit Strain and a material would have unit Strain strain if its length were doubled. (Original length = 1 ; Stretch = 1 ; Strain = f) If, then, a material could remain elastic whilst its length were doubled under a load, the stress in the material would have the value E. Example 9. A 20 mm bolt 160 mm long carries a load of 20 kN. Calculate the extension in the bolt if E = 200 000 N/mm 2 . • *U U H L ° ad 20 000 20000 « 1 XT/ 2 Stress in the bolt = -r = = -, . = 63-7 N/mm 2 Area ti ,_ rw? ji^'Z 4"(20) 2 Extension Ext Strain = \ 200000 = Extension = Orig. length 160 „ Stress E = -=. — — Strain E = 200 000 Stress 63-7 63-7 63-7 x 160 Ext Ext T60 63-7 x 160 _ 0-05 lmm 200 000 Example 10. A 20 mm steel bolt is threaded through a brass sleeve 100 mm long, 24mm bore and 32mm outside diameter. A nut and washer are 196 MECHANICAL PRINCIPLES — II put on and the nut tightened up until the brass sleeve has shortened by 0-05 mm. Calculate the extension in the bolt. E for steel = 200 000 N/mm 2 E for brass = 80 000 N/mm 2 Strain in sleeve = -^ = 0-0005 and since E =|==^: Stress = ^(Strain) = 80 000 X 0-0005 = 40N/mm 2 Strain Stress in sleeve = 40 N/mm 2 Cross-sectional area of sleeve = |(32 2 - 24 2 ) - ==(1024 - 576) 22 x 448 a „ 2 = ^ = 352 mm 2 Hence compressive load carried by sleeve = (Stress) (Area) = 40 x 352 = 14 080N This will be equal to the tension in the bolt. •1-1* 14080 14080 AA OXT/ 2 Hence stress in bolt = = ,. . ^ = 44- 8 N/mm 2 p*y 3142 „ Stress , Ci . Stress 44-8 E = -=- — — and Strain = But strain = Strain E 200 000 Extension Ext Orig Length ~ 100 - 1^L_ 44-8 ' • 100 ~ 200 000 100 x 44-8 nM ~ A Ext = 200 000 = ° 0224mm Extension of bolt = 0-0224 mm Exercises 7c 1. A bearing has to carry a load of 3000N, with a bearing pressure of 0-75N/mm 2 . I f the length of the bearing is to be made equal to twice its diameter determine its dimensions . 2. A 50mm diameter shaft runs in two bearings spaced at 2-5 m centres, the bearings each being 80mm long. Loads of 700N, 800N and 750N act on the shaft at 0-5m, l-25m and 1 -8 m respectively, from the LH bearing . Determine the load and bearing pressure at each bearing. 3. A bearing 50mm diameter, 80mm long, carries a total load of 6000N. If the shaft is rotating at 210 rev/min estimate the coefficient of friction from the expression /u — 0-032Vv . Hence determine the number of joules of work lost in friction per second, P i.e. the power lost in watts. STRESS AND STRAIN 197 4. The end thrust on a spindle is taken by a collar on a 60mm diameter portion of the shaft. If the maximum thrust is 2100N, and the bearing pressure is not to exceed 0-7 N/mm 2 determine the necessary top diameter of the collar. 5. A line of 60mm shafting runs in six bearings each 100mm long. The average load on each bearing is 6000N and the speed of the shaft is 3 1 5 rev/min . Estimate the coefficient of friction and the power being lost in friction. 6. If the Elastic Limit of a certain material were at 160 N/mm 2 , what load would a 20 mm diameter bar of the material carry without sustaining a permanent stretch? 7. Taking the ultimate stress of a mild steel to be 500N/mm 2 , calculate the load in Newtons necessary to fracture an M6 metric bolt at the root of the thread. [Take the diameter at the root of the thread as 4-5 mm .] Hence taking the efficiency of the thread at 10%, find what force at the end of a 200mm spanner will cause the bolt to fracture. [M6 thread has a pitch of 1 mm] 8. A drawbolt 20mm dia, 800mm long, is pulled up to a tension of 16 000N. Calculate the stress in the bolt, and if E = 200 000 N/mm 2 , determine the total extension. 9. An air cylinder is 140 mm diameter and the cylinder head is held on by six M 12 studs and nuts. If the nuts are tightened up to an initial tension of 1000N, calculate the stress at the root of the thread when the air pressure in the cylinder is 0-6N/mm 2 . [Take the root diameter of the M12 thread at 9-5 mm.] 10. A steel ring 20mm wide, 10mm thick and 14985 mm inside diameter, is heated up and shrunk on to a shaft 150mm diameter. If E = 200000N/mm 2 , estimate the stress and the tension in the ring. 8 Mechanical principles - III The equations of motion The distance travelled during a given time by a body moving at a certain constant speed will be given by multiplying the speed by the time, or if s = space travelled, v = velocity (or speed), and t = time, s = vt (1) It is important that the time units of v and t are coherent (e.g. if v is in metres/ second, then t must be in seconds and s will then be in metres). Acceleration Acceleration is the rate of increase of velocity. For example: if a body starts from rest with an acceleration of 1 m/s every second (i.e. 1 m/s 2 ), its velocity at the end of 1 second will be 1 m/s, at the end of 2 seconds it will be 2 m/s, and at the end of / seconds it will be t metres per second. Hence we may say for a body starting from rest with an acceleration a: Final velocity (v) after time t ; v = at If, instead of starting from rest, the body already had an initial velocity of u, then: Final velocity after time t = u + at This gives us v = u + at (2) Instead of accelerating, a body may be slowing down or decelerating. Its acceleration will then be a minus quantity, and we shall have for its final velocity: v = u — at This slowing down is generally termed retardation. A graph of velocity-time for equation (1) above is shown in Fig. 147(a) and the area under the graph is equivalent to the distance moved. If we now plot a similar graph for equation (2) it will be as Fig. 147(6). In this case, due to the acceleration, the velocity is increasing at a constant rate and the graph is a sloping line instead of a horizontal one. The distance moved will be, as before, the area under the graph (Area O ABC). ACCELERATION 199 Fig. 147 But OABC = OADC + ADB s = ut + \tat This gives us s = ut + hat 2 If u = O then the line starts at O and s = hat 2 (3) The reader is strongly advised to interpret problems of accelerated motion as far as possible with the help of a graph. Equation (3) could have been obtained as follows: For any motion the space travelled = (average velocity) (time). In Fig. 147 (b) the average velocity is at a height midway between A and B, i.e. u + \at. Multiplying this by the time t we get s = ut + hat 1 as before. Finally, we require an equation connecting v, s and a, so we must eliminate t. and In equation (2) we have v = u + at v — u t = a Substituting this value for t in equation (3) we have s = ut + hat 2 uv - u 2 , /v 2 - 2uv + u 2 \ -^r- + *\ — ? — ; 200 MECHANICAL PRINCIPLES — III Putting on the common denominator 2a 2uv — 2u 2 + v 2 — 2uv + u 1 s = s 2a u 2 2a which gives us that v 2 — u 2 = 2as (4) If the initial velocity u = (body starting from rest) then v 2 = las. In problems dealing with the motion of bodies falling under the action of gravity then, a = acceleration due to gravity = 9-81 m/s 2 This is generally signified by g instead of a, and the reader should note particularly that in using 9-81 forg the units are in metres and seconds. The units of all the other quantities in the equations must be kept in coherent units. Example 1 . A drop stamp falls freely for 6 m under the action of gravity. Find its velocity at the moment it strikes the tup. Here u = s = 6ma = 9-81 m/s 2 and v 2 = las = 2 x 9-81 x 6 = 117-72 v = Vl 17-72 = 10-85m/s Example 2. A shaping machine ram is running on a stroke of 450mm. It starts from rest, accelerates at a uniform rate until the centre of the stroke and then retards at a unifrom rate to a standstill at the end of the stroke. If the stroke occupies \\ second, find the acceleration and the maximum speed attained. This problem is best illustrated graphically, and the motion is repre- sented in Fig. 148. The ram is accelerated from O to A, its velocity increasing uniformly and the reverse process takes place from A to B. The area OAB represents the space travelled, which in this case is 450 mm = 0-45 m ACCELERATED MOTION 201 Hence 045 = £OB.AC and since OB = 1£ seconds 045 = i.HAC = |AC .-. AC = v = 0-6 m/s For half the stroke v = at 0-6 = a. \ anda = 0-8m/s 2 Hence acceleration = 0-8 m/s 2 max speed of ram = 0-6 m/s Note that the average speed of the ram = 0-3 m/s Vel. Fig. 148 Fig. 149 Example 3. A planing machine table is set for a travel of 900 mm. It starts from rest, accelerates uniformly during the first 1 50 mm, runs at a constant speed for 600 mm and then retards uniformly to rest during the last 1 50 mm of the travel. The total time taken travel the 900mm is 4 seconds. Find (a) the average speed, (b) the maximum speed, (c) the acceleration. A graph showing the motion is shown at Fig. 149. Let the maximum velocity be v and the times of acceleration, uniform speed and retardation f,, t 2 and tj, as shown. (a) The average speed will be distance 0-9 m = 0-225 m/s time ~~ 4s Since the first and last portions of the travel are identical: retardation = acceleration (a). For the first part of the stroke: v 2 = las = 0-3a (since s = 0-15 m) (1) 202 MECHANICAL PRINCIPLES — III Also, for the first and last portions of the travel v = at, = at, and t, = t, = — a and for the second s 2 = vt 2 t 2 — — (since s 2 = 600 mm = 0-6 m) .,,,,, v , 0-6 v . 2v 0-6 ,«v ..t l +t 2 + t i =- + — + - = 4 = — + — (2) a v a a v (since total time = 4 seconds) v 2 From equation (1): # = — Substituting this in equation (2): 2v x 0-3 0-6 _ V 2 V 0-6 0-6 . . — + — =4 i. V V and 4v= 1-2, v Since from (1) v 2 0-3 x 0-3 a ~ 0-3 ~ 0-3 ii=4 v = 0-3m/s 2 Exercises 8a 1. A shaft starts from rest and with uniform acceleration attains a speed of 300 rev/min in half a minute. Sketch the graph of velocity-time, and calculate (a) the acceleration in rev/min 2 ; (b) the number of revolutions made by the shaft during the period. 2. A drop stamp falls freely under the action of gravity from a height of 8 m. Calculate (a) the time of fall, and (b) the velocity at the instant it strikes the bottom block. 3. Starting from rest, a shaping machine ram, with uniform acceleration, reaches a speed of 24 m/min during 240mm of travel. Find the acceleration and the time taken. 4. A cam rotates at 180 rev/min. During 90° of its revolution it causes a plunger to rise a distance of 25 mm. Half the rise is made with uniform acceleration and the remainder with an equal retardation. Calculate the acceleration of the plunger and sketch the graph of velocity-time for it. 5. The total travel of a planer table is 2-1 m. Starting from rest, the table accelerates uniformly for \\ seconds, runs at a constant speed for 2 seconds, and retards to rest during 1| seconds. Calculate the acceleration, and the speed during the middle interval. Sketch the graph of velocity-time. 6. After the power has been shut off, a flywheel rotating at 180 rev/min slows down to rest during 90 turns. If the retardation is uniform finds its value in rev/min 2 and the time taken by the wheel in coming to rest. 7. A machine slide has an acceleration of 0-08 m/s 2 . How far will this slide travel from MASS AND WEIGHT 203 rest before reaching a speed of 24m/min, and how long will it take for this to be effected? 8. A slide starts from rest and travels a distance of l-25m with an acceleration of 0-1 m/s 2 , and then comes to rest again with a retardation of 0-05 m/s 2 . Calculate (a) the maximum speed attained, {b) the total time taken. Sketch the graph of velocity-time. Motion and force Newton's First Law of Motion states that every object remains at rest or moves with uniform velocity in a straight line until compelled by some force to act otherwise. His Second Law states that change of motion is proportional to the impressed force, and takes place in the direction in which that force acts. Before we can arrive at a quantitative relation between force and change of velocity we must consider the velocity of an object as something more than a rate of movement, since force required will depend upon the size or amount of material in the body. Mass and weight Every object is made up of a mass of material (iron, wood, stone or what- ever it might be), and this mass is constant and invariable, so long as we do not cut away from or add material to it. Due to the gravitational pull of the earth acting on this material every body exerts a downward force. This force is the weight of the body. The weight of a body is not a constant quantity because the pull of gravity varies according to the distance from the centre of the earth. For example, the weight of the same amount of mass would be about \ percent greater at the poles than at the equator. We thus have the mass of a body, being the amount of substance in it, and a constant quantity, and the weight, being the downward force caused by the gravitational pull acting on the mass. We may now define the quantity of motion of a body as being (mass) (velocity), and for the given velocity this will not vary since mass is a constant quantity. This product mv (m = mass) is called Momentum. By Newton's Second Law: Force is proportional to change of motion, i.e. Force is proportional to change of mv. But m cannot change as it is constant. .-. Force varies as m (change of v) Now change of velocity is acceleration. Hence Force varies as ma . If the units of F, m and a are chosen so that unit force causes unit 204 MECHANICAL PRINCIPLES — III acceleration on unit mass we may say F = ma, and this is the fundamental relationship between the quantities. The SI system of units is coherent. This means that the SI unit of force is the force that gives to one unit of mass one unit of acceleration. To honour the contribution of Sir Isaac Newton to Science this is called the newton. Hence F (newtons) = m (kilogrammes) X a (metres per second every second) Let us now consider the action of gravity on a mass of m kilogrammes, i.e. the weight of the body. Using F = ma a = g and F — mg The weight of a body of mass m kilogrammes is, therefore, mg newtons; g may be taken as 9-81 m/s 2 . Example 4. If the planing machine table in Example 3 has a mass of 400 kg, and the coefficient of friction at the slides is 0-1, calculate (a) the total force necessary to accelerate it, and (b) the power being taken to accelerate and overcome friction at the instant when the table has moved 75 mm from the beginning of its travel. (a) Force to overcome friction = (W)/u = (mg)fi = 400 x 9-81 x 0-1 = 392-4N Force to accelerate = ma = 400 X 0-3 = 120N Total force = force to overcome friction + force to accelerate = 392-4 + 120N = 512-4N (b) After 75 mm of movement, speed of table = -1- = 0-3 m/s Rate of doing work = (force) (speed) = 512-4 x 0-3 = 153-75W Power = 154W = 0154kW We might add that during the middle portion of the stroke no acceleration is taking place and the table has merely to be kept moving against friction (neglecting any cutting force). Here we have: frictional resistance = 392-4 N FORCE OF HAMMER BLOWS 205 Speed = 0-6 m/s Power = 3924 x 0-6 = 23544 W = 0-235 kW Force of hammer blows The force of a hammer blow may be estimated by considering the retarda- tion of the hammer as shown by the following example: Example 5. A hammer of mass 1kg and moving at 2 m/s strikes a pin and is brought to rest by driving in the pin 5 mm. Calculate the average force of the hammer blow on the pin. To find the retardation of the hammer we may use the equation v 2 — u 2 = las Here v (final velocity = u (initial velocity) = 2 m/s s = 5 mm = 0-005 m — u 2 = las I 2 a = — 1 x 0-005 Force = ma = 1 X 400 = 400 N = -400 m/s 2 Example 6. A machine slide, of mass 500 kg and moving at 24m/min, takes 1 second to come to rest after the power is shut off. Calculate the average Motional resistance assumed as a force acting against the slide. To find the retardation of the slide we may use the formula v = u + at Here v = 0: « = |£= 0-4 m/s: t = 1 second = 04 + a(l) a = -04 m/s 2 (-ve because retardation) If the average resistance is denoted by F F = ma = 500 x 04 = 200N Exercises 8b 1. A slide of mass 100kg starts from rest and accelerates uniformly to a speed of 12m/min in 2 seconds. Neglecting friction, calculate the force necessary to produce the acceleration. 2. A machine table has a mass of 60kg and the frictional resistance at the slides is 206 MECHANICAL PRINCIPLES — III equivalent to a force of 50N. If a constant force of 80N is applied to this slide, what speed will it have attained after 2 seconds? 3. If the planer table in Ex. 8a, No. 5, had a mass of 300kg, and the coefficient of friction at the slides were 0-08, what force would have been required to accelerate it during the first portion of its travel? 4. A lkg hammer head moving at l-2m/s is brought to rest by driving a pin through a distance of 10 mm. Assuming the retardation of the hammer to be constant, calculate the average force of the blow delivered to the pin. 5. Taking the mass of the cam plunger in Ex. 8a, No. 4, to be 1-2 kg, calculate the vertical force exerted by the plunger on the cam face during the portion of the lift that the plunger is accelerating. [The plunger moves upward on a vertical centre line.] 6. A shaping machine ram has a mass of 150kg and on its return stroke it starts from rest and moves through a distance of 360 mm in 0-75 s with uniform acceleration. Neglecting friction, calculate the acceleration, the accelerating force, the velocity after the ram has moved 360mm and the power necessary to move it at that instant. 7. A drop stamp of mass 100kg falls freely from a height of 6m. Calculate its final speed. It is brought to rest by compressing the metal on the bottom block through a distance of 10mm. Determine the retardation of the stamp and the average force ofthe blow delivered . 8. A machine slide of mass 50 kg is moving at 0-8m/s when the power is shut off. If the frictional resistance to its motion is equivalent to a force of 50N, how far will it travel before coming to rest? Energy There are various forms of energy (e.g. heat energy, chemical energy, electrical energy, etc) and the origin of all of them may be traced back to energy derived from the heat ofthe sun. The Law ofthe Conservation of Energy states that energy cannot be destroyed: one form may be changed into another, but we can neither create new energy nor destroy that which is in the universe. At the moment we are interested in the energy of mechanical movements, and in this connexion we may define the energy of a body as the power of overcoming resistance or of doing work. A body may possess energy by virtue of its position, e.g. a weight on a cord wound round a shaft may rotate the shaft and do work as it descends to the floor. A wound-up clock-spring possesses energy. Energy of this kind is called Strain Energy. When a body is moving, it possesses energy of a different kind called Kinetic Energy. A rotating flywheel or an oscillating machine slide are examples of moving bodies possessing kinetic energy. Since we may convert energy from one form to another without loss, we may obtain an expression for kinetic energy: Let a body of mass m (i.e. weight mg) be raised to a height h above some datum line (Fig. 150). When in this position the body will possess ENERGY 207 mgh units of potential energy, since this amount of work must be expended to lift it there. If now the body be allowed to fall, it will lose its potential energy, and gain an equal amount of kinetic energy and KE = PE = mgh We may express the K.E. in terms of v because v 2 — u 2 = las. u = 0; a = g, and s — h. :.v 2 = 2gh and Hence KE = mgh 2g v 2 mv 2 mgx Tg = ~T (6) An expression for kinetic energy in terms of m and v. The SI unit of all forms of energy is the newton metre, i.e. the joule (J). The reader is reminded that energy and work are interchangeable: to put a body in possession of a certain amount of energy requires the expenditure of an identical amount of work. m 1 1 1 F = \mg v\\ I 1 1 1 Datum L——L.UPS—1 Fig. 150 Example 7. A shaping machine ram has a mass of 200 kg and accelerates uniformly from rest, covering the first 360 mm of its travel in 0-75 s. Find the velocity at the end of this travel, and show that the kinetic energy of the ram at that point is equal to the work done in accelerating it. In Fig. 151 the motion of the ram may be represented by the lineOA, and AB represents the velocity v attained after 0-75 second. 208 MECHANICAL PRINCIPLES — III Since space travelled = Area OAB = 360 mm = 0-36 m i(v X !) = 0-36 and 5" =0-36, v = 0-96 m/s o v To find the acceleration we have v = at or a = - 096 ne / 2 0T75 = 1 - 28m/S Force = ma = 200 x 1-28 = 256 N Work done = (force) (distance) = 256 x 0-36 = 92- 16 J Energy of 200kg ram at 0-96 m/s = ^ = 200 x °' 962 = 92-16 J, as before, Example 8. Solve Example 6 by treating it from the aspect of energy. Kinetic energy of 500 kg slide at 04 m/s mv 2 _ 500 X 0-4 x 0-4 _ 2 ~ 2 The slide is stopped in 1 second from a speed of 0-4 m/s Average speed = 0-4 -~ 2 = 0-2 m/s space = (velocity) (time) =0-2x1= 0-2 m Hence 40 J of energy is dissipated over a distance of 0-2 m and if F is the force acting to do it Work = (force) (distance) 40 = F x 0-2 F = 40 -s- 0-2 = 200 N as before. Example 9. A 250 kg drop stamp falls through a height of 4 m before striking the work. If it compresses the metal and is brought to rest in a distance of 25 mm, estimate the average force of the blow. Work stored up in stamp when striking work = potential energy = mgh = 250 x 9-81 x 4 = 98 10 J This is dissipated over a distance of 25 mm = 0-025 m Work = (force) (distance) 9810 = F x 0-025 7om 392 400N r aQ25 j?x huui> CIRCULAR MOTION 209 Circular motion Many of the problems of the shop deal with rotating masses (flywheels, pulleys, etc), so that we must adapt our knowledge of motion and force to circular as well as linear movement. It is quite permissible to use the equations of motion as they stand and apply them to some point on the rotating body. In general, this will be a point on the rim should the body in question be a flywheel or pulley. It is better, however, when dealing with circular motion, to use the angular notation of quantities, since if we consider any rotating body the speed of points at different radii will vary according to their radius. The angular speed of the body, however, is constant, irrespective of the radius. Fig. 152 We have seen that if we consider an angle AOB, where AB is the arc AB struck from the centre O, and r is the radius, then = 6 in radians. r If this refers to a body having circular motion about O as centre: if arc AB = space moved (s), then - = angle (radians), or s = Or Again, if AB represents the velocity (v) of the body at radius r, then - = angular velocity (&>) or v = cor (7) If v is in m/s, and r in m, then to will be in radian per second. In the same way the relation between the linear acceleration a and the angular acceleration a of a body moving in a circle is that a — = a or a = ar r 210 MECHANICAL PRINCIPLES — III When the reader has become accustomed to these simple conversions from the linear to the angular notation, he will find the angular working is preferable to and more rational than linear working for problems of rotation . The equations of motion given on pages 198 later may be used for angular notation when it is remembered that: s = space in radians; v = velocity in rad/s a = acceleration in rad/s 2 Since 1 revolution = In radian we may convert rev/min to rad/s by dividing by 60 and multiplying by 2n, N r i i.e. (o (rad/s) = -tk^ 71 - l N = rev/minJ oU Example 10. A pulley revolving at 220 rev/min comes to rest with uniform retardation in 50 turns. Calculate the angular retardation and the time taken . We may obtain the time very simply without using radians, for if the retardation is uniform, the velocity-time graph is a straight line and the average velocity = —=- = 1 10 rev/min The pulley therefore comes to rest in 50 turns at a mean speed of 1 10 rev/min, i.e. in fYfc min = ffi x 60 = 27 sec. Using the equation cv 2 = oj x + at U = -pfT-Mi + a. 27. oU 220 X 2n „ _, ,. , a = - ~zr ^pt = °' 853 rad/s 2 Ol) X 2/ Example 11. A gas engine which normally runs at 500 rev/min takes 20 s to accelerate its flywheels to their full speed. Calculate the angular acceleration and the number of revolutions made by the engine. (Assume constant acceleration.) We may obtain the number of revolutions from the average speed as before, since average speed = -=— = 250 rev/min 250 Using s = vt we have s = -^ X 20 = 83-3 revolutions. oU ACCELERATING TORQUE 211 To find the acceleration we may use a> 2 = a> l + at 500 on -Zrr-2 71 = + a.20 ol) 500 X 2.T a = -2- = 2-63 rad/s 60 x 20 6 Exercises 8c 1. Convert a speed of 175rev/min to rad/s, and llOrad/s to rev/min. 2. A pulley has an acceleration of 4 rad/s 2 . How long will it take to reach a speed of 250 rev/min? 3. A shaft retards from a speed of 500 rev/min to rest in 3| second. Calculate its retarda- tion in rad/s 2 . 4. Calculate the kinetic energy of a car of mass 1400 kg and travelling at 54km/h. If this is brought to rest by the brakes in 30m, find the average force of resistance exerted. 5. Solve Ex. Sb, No. 7, from a consideration of energy. 6. A machine table of mass 150 kg is moving at 30m/min. Calculate its kinetic energy. What force applied to it will bring it to rest in 600mm? (Neglect friction.) 7. Find the energy stored in a 4kg hammer-head moving at 2m/s. If this hammer is brought to rest by compressing the metal under it through a distance of 25 mm, determine the average force of the blow. Accelerating torque When a torque acts on a rotating body or on a body capable of rotation it causes angular acceleration. In Fig. 153 let the small mass m be rotating about the centre O at radius r and let the force F act on it. Then F = ma, where a = linear acceleration of m. But torque T = Fr, and angular accel a = -, i.e. a = ar. Hence F = ma = mar and T — Fr = mar 2 . But a rotating body is made up of many small masses m, each situated at its own particular radius. The sum of all these small masses we will call M, the total mass of the body. If this mass could be concentrated at a single radius which we will call k, we could write for the whole rotating body: T = Mk 2 a . Actually it is possible to determine the value of a radius at which the whole of the mass of a body may be assumed as concentrated, and this radius is called the Radius of Gyration, being denoted by k. 212 MECHANICAL PRINCIPLES We may thus write: Accel torque (T) = Mk 2 a (8) For a flywheel having a rim heavy in proportion to the rest of the wheel, k may be assumed as the mean radius of the rim. For a plain disc of radius r, k = 0-707 r. We might caution the reader again regarding the units of the above expression. If T is in newton metres (Nm), M will be in kilogrammes, k will be in metres and a will be in rad/s 2 . S/ I m F O l s \ / If 60*0 25 * 40N 160N Fig. 153 Fig. 154 Example 12. A flywheel of mass 550kg has a heavy rim, the inside and outside radii of which are 450 mm and 550mm respectively. When this wheel is rotating at 105 rev/min a brake is pressed against the rim with a radial force of 160N. If the coefficient of friction between the brake and the wheel is 0-25, calculate how long the wheel will take to come to rest and how many revolutions it will make in doing so. The tangential force at the rim of the wheel tending to stop it is 160 x fx. = 160 x 0-25 = 40N Since this acts at a radius of 550 mm = 0-55 m, the retarding torque will be 40 x 0-55 = 22Nm. As the rim is heavy in comparison with the rest of the wheel, we will take the Radius of Gyration at the mean rim radius, i.e. at 500mm radius (0-5 m). ACCELERATING TORQUE 213 Using the torque equation T = Mk 2 a we have 22 22 = 550 x 0-5 X 0-5 x a from which a = 5Q Q - Q - = 0-16rad/s 2 We may now use the equation co 2 = w, + at to find the time to stop the wheel since n 105 n ^ and a = —01 6 (retardation) Hence = -^-2* -0-16* i.e. In — 0-16? 60 t = tr n X n %* = 68 ' 8 seconds 60 x 7 x 0-16 Since the average speed of the wheel in coming to rest is 52-5 rev/ min it will make 52-5 X -pk~ in 68-8 sec, i.e . 60-2 re v. 6U Example 13. A pulley is in the form of a cast-iron disc 500mm x 80mm thick. It is driven from a source of power which exerts a constant torque of ION m. How long will this pulley take to attain to a speed of 210rev/min when started from rest? Take the density of the cast iron to be 7280 kg/m 3 . Mass of pulley = j DH X density = 2| x . 5 2 x . 08 x 7280 kg = 1144 kg Taking the radius of gyration as 0-707 x outside radius, we have k = 0-707 X 0-25 = 0-1 72 m The applied torque is lONm and applying the torque equation T = Mk 2 a, 10 = 114-4 X 0-172 2 X a <x = ttt-a nmi = 2-8rad/s 2 H4-4 x 0-172 2 co = 2l0rev/min = 44rad/s 44 1 ~ S3 15-7 seconds 214 MECHANICAL PRINCIPLES — III Example 14. A steel level 300mm long is of rectangular cross-section 40 mm x 20 mm, and is pivoted at its centre . One end bears on a cam which causes it to swing through a length of 25 mm, during the time that the cam makes \ revolution at 180rev/min. If half the swing of the lever is made with constant acceleration, and the other half with constant retardation, estimate the torque required to accelerate the lever. Ik for a lever pivoted at the centre = -r-approx(L = length of lever).] Take the density of steel as 7840 kg/m 3 The problem is shown in Fig. 155. Fig. 155 Approximate mass of lever = 0-3 x 0-04 x 0-02 x 7840 = 1-88 kg. The lever is accelerated through an arc 25 mm long on a 150 mm radius, i.e. through an angle of £ radian. This takes place during £ turn of a cam revolving at 180 rev/min, i.e. \ rev at 3 rev/s = £ X j = 27 second Since = <ot + \at 2 = \at 2 when co = If) 1 v i .-. 9 = \at\ and a = =g- = ^jkr = 192 rad/s 2 For the lever we have radius of gyration k = -pr = — yy — = 25 mm = 0-025 m, and mass m = 1-88 kg. Hence applying the torque equation T = mk 2 a = 1-88 x (0-025) 2 X 192 = 0-225Nm This means that with the cam end of the lever being accelerated upward a torque of the above amount would have to be applied by the cam to the THE ENERGY OF ROTATING BODIES 215 end of the lever. With the cam end falling, this torque would have to be applied to the lever by a spring or other means to prevent the roller from leaving the cam. The above consideration is, of course, relative to the lever only, and neglects the effect of attachments to its free end. The energy of rotating bodies If we continue our consideration of a rotating body as having all its mass concentrated at the radius of gyration (k), if the body is rotating at a speed of a> radians/second the linear speed in metres/second of a point at radius k will be v = o) k, if k is in metres. mv 2 Since the kinetic energy of a body = -^-; for a rotating body it will be m(a)k) 2 newton metres Hence KE = — -= — joules Rotating flywheels are used on presses and other machines for the purpose of storing a reserve of energy, so that when a sudden large out- put of work is required the machine will not stall, or undergo an undue slowing up in speed. Example 1 5 . Calculate the energy stored up in a solid disc of cast iron, 500 mm dia by 80 mm thick, when rotating at 180 rev/min. If this wheel is on a press, what will be its speed after a hole has been blanked if the blanking pressure is 100 kN and its duration extends over 5 mm. From Example 13 the mass of this wheel is 114-4 kg and its radius of gyration, 0-1 72 m. 180 The angular speed o> = -^r- x In = 6;rrad/s oil Putting these in the expression for Kinetic Energy „„ mk 2 (o 2 114-4 x 0-1 72 2 x (6tt) 2 KE = - T - = 2 Energy stored at 180 rev/min = 601-5 J 216 MECHANICAL PRINCIPLES — III For a pressure of 100 kN extending over 5 mm, the work done = 100 000 x 0-005 = 500J This energy must be given up by the flywheel and its energy afterwards will be 601-5 - 500 = 101-5 J If oj x is the speed of the wheel after giving up the energy: re-applying the expression for kinetic energy Tm , mk*a>* 2 2 x 101-5 203 101-5 = — - — : u> 1 — 2 ' rnk 2 114-4 X 0-172 2 = 60 (jo = ^60 = 7-75 rad/s co 7-75 rev/s = x- = -= — In In 7-75 X 60 232-5 _, , . rev/m = = = = 74 rev/min 2.71 71 Hence speed of wheel after blanking hole = 74 rev/min Example 16. Solve Example 12 from a consideration of energy and work done. At 105 rev/min the work stored up in the flywheel = Energy = — = — 550 X 0-5 X 0-5 X 11 X 11 2 = 8320 J The tangential force at the brake block = 40 N, and to dissipate 8320 J 8320 of energy this must travel —tpt- = 208 m Revs of wheel for 208 m on its circumference 208 208 , A « , , 60-2 rev, as before Circum n x 1-1 Average speed = 52-5 rev/min Hence time to stop = -py-^ X 60 min = 68-8 s Example 17. A fly-press has two 150mm diameter spherical cast-iron balls spaced at the ends of the arm, each ball being at 500 mm radius. At THE ENERGY OF ROTATING BODIES 217 Fig. 156 what speed must the arm rotate if the punch is to be just capable of penetrating a thickness of 2- 5 mm under a constant pressure of 20 000 N? Take the density of cast iron to be 7280 kg/m 3 . The reader is no doubt acquainted with the fly-press, a sketch of which is shown in Fig. 156. mass of a ball = volume x density = n * ai53 x 7280= 12.9kg Energy stored in 2 balls at a speed of co rad/s. = 2 pM ) = mk2(o2 = 12-9 X 0-5 X 0-5 X <o 2 = 3.225a; 2 This must be equal to the work output required at the punch, i.e. 20 000N acting for a distance of 2- 5 mm = 0-0025 m = 20 000 x 00025 = 50J Hence 3-225w 2 = 50, = Vl5-5 50 3-225 = 15-5, 3-94 3-94 rad/s = ^P = 0-627 rev/s 218 MECHANICAL PRINCIPLES — HI Exercises 8d 1. A CI flywheel rim is .1 m outside, 0-8m inside diameter and 01m wide. Taking the radius of gyration as the mean radius, calculate the energy stored up when the speed is 315rev/min [Density of CI = 7280 kg/m 3 ]. 2. If the flywheel in Ex. No. 1 is attached to a press and a punching operation causes the speed of the wheel to drop to 262-5 rev/min, calculate the energy absorbed by the operation. 3. A CI flywheel, rim 14m outside diameter, lm inside, and 01m wide is revolving at 105 rev/min when two brake pads are pressed against opposite sides of its rim. If the coefficient of friction at the brakes is 04, with what force must they be pressed against the rim to bring the wheel to rest in 30 revolutions? 4. A motor develops a constant torque of 32Nm. It drives a shafting and pulleys of mass 400 kg, and having a radius of gyration = 200 mm. Calculate the time taken for the motor to accelerate the shaft from rest to a speed of 3 1 5 rev/min. 5. A flywheel is in the form of a solid cast-iron disc 0-6 m diameter, 80mm thick, and is carried on a 50mm diameter shaft. When the wheel is revolving at 315 rev/min it is dis- engaged from its drive and comes to rest in 75 seconds . If the retarding torque is due entirely to friction at the bearings, calculate the tangential frictional force at the surface of the 50mm shaft upon which the wheel is mounted. 6. A press has a flywheel of mass 400 kg and radius of gyration = 0-707 m. The press is shearing metal bars 40mm x 20mm, for which the ultimate shear strength is 400N/mm 2 . When a bar is being sheared the full load comes on and remains constant whilst the shear blade moves 5 mm, after which the load drops to zero. If the above flywheel is rotating at 52-5 rev/min when a bar is sheared, calculate the reduction in speed caused by the work absorbed. 7. A fly-press has two masses each of 5 kg attached to opposite ends of the arm at 400mm radius. When the arm is rotating at 1 rev/s a punch in theram makes contact with a pin and drives it a distance of 5mm into a hole. Calculate the average pressure exerted on the pin by the punch. 8. A cylindrical mixing tank mass 160kg, inside diameter lm, radius of gyration 250 mm, is rotating at 105 rev/min when 20 kg of material is tipped into it. If the material immediately moves to the sides of the tank and arranges itself at a mean radius of 400 mm, estimate the momentary reduction in speed caused by the extra mass. Heat and Heat Energy Temperature The temperature of a substance is no indication of the amount of heat it contains, but merely a measure of its "hotness level." There are two scales of temperature used in this country; the Celsius and the Kelvin. (The Celsius scale is a more modern* name for the Centi- grade scale, but the latter name is likely to persist for some time after the adoption of the SI conventions.) The freezing-boiling point limits of water are shown in Fig. 157. AMOUNT OF HEAT 219 373 273 100" Boiling Freezing Kelvin Celsius Fig. 157 The degrees of temperature on these systems are generally denoted °C and K respectively. The chief calculation concerning these is the conver- sion from one to the other. It will be noted that a very simple relation- ship exists, i.e. temperature in Kelvin = temperature in Celsius + 273. For example 20°C = 273 + 20 = 293K and 353 K = 353 - 273 = 80°C (The reader will note that the word "degree" is not used when referring to the Kelvin scale.) Furthermore, 1°C of temperature rise = 1 K of temperature rise. The symbol relates to the kelvin scale of absolute temperature and goes down to absolute zero (0) of temperature. It is employed mostly in scientific and thermodynamic work. Amount of heat Heat is a form of energy, and hence the unit for a quantity of heat is the same unit as for a quantity of energy, i.e. the joule. For large quantities we occasionally refer to kilojoules and megajoules.The amount of energy needed to raise unit mass of a substance a unit increase of temperature is called the specific heat capacity of that substance. As an example, the specific heat capacity of water is about 4200J/kg°C. Water has the greatest specific heat capacity of any substance, so that all other sub- stances have a specific heat capacity of less than 4200 J/kg°C. 220 MECHANICAL PRINCIPLES — III Table of Specific Heat Capacities Specific Heat Substance Capacity in J/kg °C Water 4200 Alcohol (absolute) 2930 Olive oil 1300 Petroleum 2140 Turpentine 2000 Copper 400 Cast iron 540 Aluminium 900 Lead 1300 Steel 480 Brass 395 Oak 2400 Stone generally 880 Air (at constant pressure) 1000 When a substance is heated up the heat transferred to it will be given by (Mass of substance)(Rise in temperature)(Specific heat capacity) = mc (T 2 - T,) [m = mass;c = sp.ht. capacity; T x and T x = Temp limits.] Heat calculations Due to the fact that heat escapes so easily and so rapidly by conduction, convection and radiation, the results of heat and temperature calculations are not to be relied upon. It is possible, however, to obtain an approximate idea of the state of conditions, and the work involves important principles in which we cannot have too much practice. Example 19. A piece of steel of mass 5kg is heated up to 700° C and immersed in a tank 400mm square containing 250mm of water at 15°C. The steel is left until temperature conditions become steady. Assuming that 10% of the heat is lost, estimate the final temperature. When the hot steel is placed in the water it will lose heat to the water, and if no heat were lost externally the heat lost by the steel would be HEAT CALCULATIONS 221 equal to the heat gained by the water. As it is, 90% of the heat lost by the steel is gained by the water. Let 7 be the final steady temperature. Heat lost by steel = (mass)(sp. ht. cap.)(fall in temp) = 5 X 480(700 - 7) The mass of 1 cubic metre of water = 1000 kg, so the mass contained in a tank 04 m square when the depth is 0-25 m = 0-4 x 0-4 x 0-25 x 1000 = 40kg Heat gained by water = 40 x 4200(7 - 15) (2) Since heat gained by water = i%(heat lost by steel) we have 168000(7 - 15) = U5 X 480(700 - 7)] (3) Multiplying out the brackets we have: 168 0007- 2 520 000 = 1 512 000 - 2 1607 From which 170 1607 = 4 032 000 ««h t 4Q32Q0Q ^ 7 or and T= 170160 = 2H1C Example 20. A piece of steel 1 kg in mass is taken from a furnace and placed in 2 litres of water. Conditions are made such that the minimum of heat escapes. If the initial temperature of the water was 20°C and the final steady temperature of steel and water 52° C, estimate the temperature of the steel when taken from the furnace. If we assume no loss of heat we have Heat gained by water = Heat lost by steel. Let 7 = furnace temperature Mass of 2 litres of water = 2 kg Heat gained by water = 2 X 4200(52 - 20) = 8400 x 32 = 268800 J Heat lost by steel = (mass)(sp. ht. cap.)(7 — 52) = 1 x 480(7 - 52) *= 480T - 24 960 Equating: 4807- 24 960 = 268 800 4807 = 293760 293760 1 480 21±±- Actually this would be a low estimate as some heat would be lost. 222 MECHANICAL PRINCIPLES — III Exercises 8e 1. The dimensions of a workshop are 40m x 16m x 12-5 m and the ventilation system is such that the air in the shop is changed twice per hour. If the temperature in the shop is 20°C and the outside temperature 12°C, calculate the heat lost per hour by the shop due to the two air changes. [Sp. ht. capacity of air = 1000 J/kg °C, density of air = l-3kg/m 3 .] Give the answer in megajoules. 2. A piece of steel mass 5 kg is taken from a furnace at 700° C and plunged into 8 litres (8kg) of water at 15°C. If 10% of the heat is lost, estimate the temperature of the water when steady conditions have been reached. [Sp. ht. capacity of steel = 480J/kg °C.] 3. Water flows through a gas-heated boiler at the rate of 12 litres per minute, and its temperature is raised from 20°C to 70°C by the boiler. If the efficiency of the boiler is 70% and the gas used yields 18MJ/m 3 when burned, calculate the gas consumption of the boiler in cubic metres per hour. 4. A piece of steel of mass 1-25 kg is taken from a furnace and quickly transferred into a tank containing 3 litres of water. The initial temperature of the water was 15°C, and the final steady temperature was 50° C. Assume that no heat escaped during the process, and estimate the temperature of the steel when taken from the furnace. 5. An oil-cooling arrangement consists of a nest of tubes through which the oil flows, the tubes being subjected to a circulation of cooling water on their exterior. The oil flows through the tubes at the rate of 18 litres/min, its entering temperature is 80°C, and it is desired to cool this down to 30°C. Estimate what flow of water will be necessary if it enters at 18°C and leaves at 28°C. [l litre oil = 0-9 kg, sp. ht. cap. of oil = 2000 J/kg °C.] 6. In a certain locality the cost of gas for industrial heating was 4p per cubic metre, and the cost of coal £1200 per tonne (= 1000kg). The heating value of the gas was 18MJ/m 3 and of the coal 25MJ/kg. Assuming an efficiency of application of 90% for the gas and 60% for the coal, compare the relative costs of heating by gas and coal. Heat energy Heat is a form of energy, and when mechanical work is dissipated by fric- tion it is converted into heat. Most of our mechanical energy is derived from heat by converting it through some form of heat engine . The relation between heat and mechanical energy was at one time called Joule's Equi- valent, after the famous scientist Joule. The use of Joule's Equivalent is not required when using SI units, as heat and other forms of energy use the same unit. Example 21. If lkg of coal when burned yields 25 MJ, estimate the mass of coal required per hour to generate 100 kW if the overall efficiency is 10%. We have that — — 2— = Efficiency = -r^. Input 100 HEAT ENERGY 223 Output = lOOkW, Input = lOOkW x -^ = lOOOkW 1000 kW = 1000 000 W = 1000 000 J/s = 10 6 J/s Heat energy required per hour = 10 6 X 3600 J Heat energy in 1 kg of coal = 25MJ = 25 X 10 6 J Mass required = — „$ x ine = 144 kg Example 22. A machining operation is cooled by soluble oil having a specific heat capacity of 3000J/kg°C, flowing at the rate of 601itres/min. If 10 kW is being absorbed at the cutting point and 80% of the heat generated is taken away by the cooling oil, calculate its rise in tempera- ture, tl litre of coolant = 0-9 kg]. Assuming all the power to be dissipated as heat at the cutting point, we have: Heat generated per minute at the cutting point = lOkW for one minute = 10 OOOJ/s x 60s = 600 000J Heat taken away by coolant = & x 600 000 J = 480 000 J Heat taken up by coolant = (mass)(sp. ht. cap.)(temp rise) = (60 X 0-9) (3000) (temp rise) joules Equating these we have: (60 x 0-9 x 3000) temp rise = 480 000 J _ . 480 000 ~ Q . or Temp - nse = 60 x 0-9 x 3000 =!&-£- Example 23. The surface speed of a grinding wheel is 1200m/min. The end of a steel bar of 40mm square cross-section is pressed against the wheel for ± min, with a force of 100N. If the coefficient of friction between the steel and the wheel is 0-4, and if half of the heat generated is absorbed by the steel, being momentarily confined to a depth of 2-5 mm, estimate the temperature rise at the surface in contact with the wheel face. If the coefficient of friction is 0-4 and the radial pressure 100N, the tangential resistance will be 100 x 0-4 = 40N and the work dissipated per second = 40N x 12 ^ m = 800Nm = 800 J. .*. Work dissipated in 15 s = 12 000 J Heat input from work = \ x 12 000 J = 6000 J 224 MECHANICAL PRINCIPLES — III Taking the density of steel as 7840 kg/m 3 , the layer of metal to which this heat is assumed to be confined = volume x density. = 0-04 x 0-04 x 0-0025 x 7840 = 0-031 36 kg Heat given = (mass)(sp. ht. capacity)(temp rise) 6000 = (0-031 36)(480)(temp rise) Temp rise = 480 x 6 °0°031 36 = ^C Exercises 8f 1. A cutting operation is absorbing 2kW at the tool point. Calculate the heat being generated in kilojoules per minute. If this operation is being cooled by oil flowing at the rate of 10 litres per min, and if 90% of the heat is taken away by the oil, calculate its rise in temperature, [l litre oil = 0-9 kg; sp. ht. cap. of oil = 2500J/kg°C.] 2. Estimate the gas consumption of an engine when it is developing 12 kW with a thermal efficiency of 20%. Take the heat value of the gas used at 18MJ/m 3 , and express the answer in cubic metres of gas per hour. 3. A bearing is 140mm diameter and runs at 500rev/min. It carries a load of 20 kN and the coefficient of friction is 0-06. Calculate the work spent per minute in friction. If the bearing is to be cooled by a circulation of oil [sp. ht. cap. = 1600J/kg°C], and if the temperature rise of the oil is not to exceed 8°C, what mass of oil must flow through the bearing per minute? 4. A flywheel of mass 250 kg and having a radius of gyration of 400 mm is rotating at 210rev/min, when it is brought to rest by a brake. Calculate the heat generated at the brake. If the brake shoe is cast iron of mass 4kg, and it absorbs half the heat generated, estimate its rise in temperature. [Spec. ht. cap. of CI. = 540J/kg°C.] 5. A grinding operation is absorbing 5kW at the wheel. Calculate the rate of flow required for the cooling water if all the heat generated is taken away by the water and its rise in temperature is not to exceed 4°C. Give the answer in litres/min. .a a t> a 8 CM « E ,-v o .a r j- ~ — ' c CO t« S • -g S S 2 § « n o ^ ■« ° o CZ2 Q vO d + J3 oo V, ^ CM CO CO CO Ov Vl Vl — 1 ON CM © 00 ON © O CM © Q CM CO ^ 3 + <N ON 0O ON VI CM O0 r- r- ro — OS O r- vi co — CM CO — i ON ON v> vi v© VO d + /l 0O N — — <N CM CM ro ro CO © Vi g + >* OS VI CM cm co CM 1 ^- V) ON V) OO so ON r- VO C d + © CM Vi r~ O CM co CM CM CO g + o\ co oo — <N CM en ro ON co v> CM Vi s VO d + — — CM CM CM co co ■«t - 1 -1- O CM vi oo CM Vl CM 00 CM CO co -1 i On — < CO vO Ov CM CM VI CM ON CM o o o o o o O O 60 -J ' *■ r- © — . — CM v> CM On CM ■CO ON CO 5 v> vo r-- ON © CM Tt VI r- i- 1 i OO ^ - CM CO Tf o VI s r-~ co 00 NO On CO vo © — — CM v> CM o ro VO CO CO © Vi On U -J i 3 £S CM CO ON Vl V) 00 Vi CM -1 i VI CM O O VI O vo CM Vl 00 8 O 00 O ON ON CM ■* O OO © CM CM o vO CM VI © CO Vi Vi CO -1 i O O Vi ■t vi vo O 00 8 O CM V) 2; R o -1 i O ViOOOOQOQQOOOO O r- ©TfoooNCOTfONOVivooocovi r~ © ViOOOQOOOQOOQQ O So ON-^CMO^vir^ooo-Hcn^vo oo 13 a ca •3 « 6. a ° "g 73 CO OOOOOVOQOQOOOV. o 5 X ^HpriTj-VlvOOOQCM^tVOOOOCM v> £ — -» t-m — .— -C CM CM CM u > O pvo ©00O©vovi©Q©OOO© vi 5 X — Jo^VivOOOOCMtJ-VOOO© cm 225 Appendix II ISO Standard H Holes (Hole Basis) Tolerance Limits for Selected Holes. (Selected from BS 4500: 1969) (Unit = 0001 mm) Nominal Sizes H7 H8 H9 Hll Over Up to and includ- ing UL LL UL LL UL LL UL LL mm 6 10 18 30 50 80 120 180 mm 10 18 30 50 80 120 180 250 15 18 21 25 30 35 40 46 22 27 33 39 46 54 63 72 36 43 52 62 74 87 100 115 90 110 130 160 190 220 250 290 UL = Upper Limit LL = Lower Limit Appendix III British Standard H Hole [Limits for H6 to Hll over the Range 6mm to 180mm Abstracted from BS 1916 (1953)1 Nominal Size High Limit (unit + 0001mm) Low Limit over to H6 H7 H8 H9 H10 Hll H6toHll 6 10 9 15 22 36 58 90 10 18 11 18 27 43 70 110 18 30 13 21 33 52 84 130 30 50 16 25 39 62 100 160 50 80 19 30 46 74 120 190 80 120 22 35 54 87 140 220 120 180 25 40 63 100 160 250 226 ? E © © £ J BO C c o o. <u o U •a c ? 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CM CM CM CO co ««r ^r •* LA IA CO IO<OP>P> oo oo oo on en o CM co o co cn oo oo on on •— •— *"" •" •" « — «" *~ *"" *~ * — " — 1— r— T— r— * — OO LA cm en CO PI ON * ^^ oo LA cm on CO (OOP-* r- oo in cm en oo «— IA cn cm co cs «• r«. LA oo cm co cn CO r> ^-coo CM la on co co (OP« o* i— O on on oo oo r» co CO LA **P) N CM r- r- O CD CO OO P»* P** CO la on CM CO cp •Cf CO CM CO O « OON CO o * oo cm in CO CO P*» ^ LA Ia o> co n o o a ^ ,2- CM CM CM CO CO •* -^•Tinin cb co co p>» r*» P>« oo oo en on en oo oo on on •— «— ' t— ^ 1— • — *~ *- i— .— ,— * — •— i— i— *— "~ »— *^ """ *"~ "~ * «— OO LA CM en co co o r» ^- r- o> co CO OP>* -- oo la cm en co CO on co co o oo CM CO an co r» C3 •» OO CM LA en co co o ^ > co i- m en CM CO O CON n n co IA LA ^ «Cf CO CM CM 000)0) oo p>p> <o m LA ^ ^" CO CM on C3 on on oo o ** °P CM CO cp ^ OO CM CO C3 «3- 00 t— LA CO co p-» t— in en CO P*» v— LA ISCONtO gVj o a ^ ^. CM CM CM CO CO ««r ^ •CT LA LA LA CO CO P> P> r-« oo oo en on en oo oo on on •— *~ *~ *~ * — *~ *~ »~ * — r~ r— r- «— "" *"■* *^ *^ *"" * — W— 9— 9— *— on ^> v oo IA CM on CO CO C3 P- ^> i— OO LA CM en co co o p^ *r r— 0O LA CM co co C3 N on CO CO O * p*. «— LA on CM CO o<<)r»r- ^> oo cm co en CO N C3 «*• OO « co cm m CO in lA ^ CO CO CM CM «— o o en oo co r~ N<oinu) «ef CO CO CM ^ CO O CO OO cp "**■ °P CM CO cp ^ T* T IA cn CO P*. r— LA CO COP> r- in en CO N O «Cf , <- CO CM CO o a o ^L .1. CM CM CM CO CO CO «#■ ^P LA LA in cb <o p> p> r-» oo oo cn co on oo oo on on *~ •""" * — .— •—»— * — " — *— T— T— w— *" »— •"" ""^ *■" *~ in o n * «— oo LA CM OJ CO co C3 P» «*■ .— OO IA cm an co co o p". * w- oo cm on co co O co n 1— •* OO cm la cn co t>» C3 ^- OO •— in en cm co o ^ - N ^1000 CM LA OO CM CO O on oo oo n CO «9in<t * co CO CM «— .— o on on oo oo r» co co in * * 01 »— «— o cp co n y— LA on nisp- LA oo CO P*. «— LA CO CM op o * CO CM CO O * OO «eoN<o C3 O O ,1. ,L. ,1. CM CM CO CO CO <r -t in in in cocph-p- r^> oo oo co cn en oo oo on en CO O n * ^ OO LA CM an co (OOP" * ^ oo ia cm on CO ioor>-f ,— oo to cm on co «er oo ^ LA on n co a CO r>» •— in co cm CO en co p» o * ao cm m on CO ia on co co O LA * * co CM CM »- ^~ C3 en CDOOP-P. CO miR* « CO (MNi-O o eo r^r» <o CO CO N w— IA an co r» »- LA op CM CO O ^ op CM CO O * op cscoo<cr oo con <- in on O O ,L ,Z. ,1. CM CM CO CO co •r >r inin in co cb p> p> r-« oo oo co cn on oo oo on on on eo cr» n * •— oo ia cm en CO CO or««t r- ao ia cm en co CO o P- * •— ia cm on CO CO LA oo CM CO on pi r» o ^~ oo cm la en co CO o * P»- «— LA en cm co o CO loonr* •— O o on oo oo r~ r-. CO LA in «» co co CM N t- O O an COCONN CO •9- "OP CO CM CM co n T "* CO CM co cp ^ op NI004 OO CM CO Cp * p~ V— LA CO CO N *T* T" *T *° on O O vi. ,1. yl. CM CM CO CO CO -r <t in in in cocbr*p- r-» oo oo oo on en oo oo on on CO CO CO o n ^ > t— co lo CM en CO CO CD P~ Tf I— OO LA CM en COCOON * r— OO IA CM on la on CO CO C3 <* r~ y- LA co CM CO O CO p* .- * COCM m COCO N O *«■ n o * eo n co CO LA LA <* co n CM i- OOO) oo COP- CO CO IA •or * CO CO CM 000)03 oo CM CO o >* OO cm co a "«r CO CM CO O CO p~ r— LA en CO r~ p-iooxo r-» co n cp * CO o o ,1. ,". ,J_ CM CM CO CO CO ^P ^ LA LA in co co co p- p~ oo oo co co o» oo oo en on on cm on co co C3 r~ ^- .- CO IA cm cn co co o r~ * »— oo in CM CO CO CO o n * «— oo IA CO CO co r-» «r oon LA CD n iso-r oo i— la en cm co eo co p~ »— IA rs r-U)fiO CM CO CM CM r— omen OO p~ p-io co m ^ I * CO CM CM «— co en en eo CO CO IA* *CT * CM CO cp •* co CM LA cn CO p~ i- menn p^. t— IA CO CO p~ r- m CO CM co NU>0*t oo O O vl. ^~ r^ CM CM CM CO CO ^ -i- -<• in in coibcsrs p» oo oo oo on en oo oo en en CO oooa O O O <=> a o ooa o o oooo o o o o a o oooo O .— CM CO «-»■ IA co N oo an o 1— CM CO ^~ LA co r~ oo co o 1- P4CO * LA CON COO) e ,- Oi 00 CO in •* CM «- O o o o o o o o o oo m «- CO in CM cn CD CO o o o o o o o o o o o CM CM CM V — T— T— O O O o O CO o ** CO CM CO O •* op cp cm cp o «t CO CM to p •* 00 CM ^ r« CM P^ CO 00 ^t 6 •* T^- CM CM CO CO CO •* ^r in in 6 6 O CM in r- o cm in r*» o CM CM CM CM CM CM CM CM CM CM in r^ «- *~ »- «- CM CM CM CM CO in in in in in n cm in n CM in r> CO CD CM CO in ^-> N CO cn in o CO T— r^ co 00 •* in n cn o CM CO LO CO 00 00 00 00 en cn cn CO cn cn O *- CM o <z> o 6 6 6 6 6 6 «- CM CO tJ- in CD N 00 en «- »- ^» ■nU r H** "H «p> m|2 =P * op ■oJJ «- CM CO -tf in co r-» oo cn t- JI CM 00 CO in Tt cm cn oo (O lO « CM cn oo co in <-* CM «— <J> 00 in cm cn co CO CO CO O r» -el- r- CO CO o r- •* t- 00 in i- r- ^ *- o o o cn en en oo oo CO •* ** •<* CO CO CO CM CM CM CO r^ ^- in en co r-. o -* CO CM CD o n «- in en CO r* «— in cn o o T- «- 1— CM CM CO CO CO xt Tt in co n n n CO CO cn cn cn in in in LO in in in in in in in cm in r*» CM in !**• CM in r>- cm in r- cm in N CM m CO CM 00 in *— r^ CO CD CN 00 in «- r^ co CO CM co in t- N in *- CD CM 00 CO cn in o CO * — r* co oo •* in i CO CM CO CO co oo en T— CM ** in r^ cn o CM co in co cc »- CO «t CD N cn CM CM CM CO CO CO CO CO CO •3- •* •*•*•* •* in in in in in in in 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 OOO 6 6 6 6 X Hs $ -P # IS Z "|- % 2p * H- % ?£ nC -h # * X *p * =B 229 Appendix VI The Trigonometrical Addition Formulae It is sometimes necessary to express the trigonometrical ratios of the sum or difference of two angles in terms of their individual ratios. The condi- tions are shown in Fig. 1 58 and the formulae are quoted without proof. If the reader wishes to pursue the proof he can do so by consulting any trigonometrical textbook. (a) (b) Fig. 158 In Fig. 158 (a), sin (A + B) = sin A cos B + cos A sin B cos (A + B) = cos A cos B — sin A sin B tan A + tanB 1 - tan A tan B tan (A + B) = In Fig. 158(6), sin (A — B) = sin A cos B — cos A sin B cos (A — B) = cos A cos B + sin A sin B , . _,. tan A — tan B tan (A — B) = - t r— — ^ v 1 + tanA tanB By writing B = A in (A + B) we get: sin 2 A =2 sin A cos A cos 2 A = cos 2 A — sin 2 A = 2 cos 2 A — 1 = 1 — 2 sin 2 A (since tan 2 A cos 2 A + sin 2 A = 1) 2 tanA 1 - tan 2 A 230 Appendix VII Continued Fractions A continued fraction is a series of fractions derived from a single compli- cated fraction, each fraction in the series approaching more nearly to the value of the fraction from which the series was derived. If during the course of our work we required to obtain a gear ratio 131 (e.g. for cutting an obscure screw pitch) of, say, -r—-, we could only do it by having gears containing 131, and 353 teeth, because neither of the numbers has any factors. To cut such gears for the job would be out of the question, as in the first place they would be too large to fit on to the machine, and the cost of making them, if they were only required once, would be prohibitive. By applying the method of continued fractions to such a problem, it would probably be possible to obtain a ratio in a usable form, and having a value so close to the original, that under the circumstances it would be acceptable. For the purpose of explaining the method we will convert the above ratio to a continued fraction: Example 18. Put the numbers down as a division sum only divide the numerator into the denominator Now divide this into the original divisor Divide into the last divisor Divide into the last divisor (and so on every time) 131)353(2 262 91)131(1 9J_ 40)91(2 80 11)40(3 33 7)11(1 7 4)7(1 4 3)4(1 3 1)3(3 3 231 232 APPENDIX VII The quotients obtained from the continued division are 2, 1,2,3, 1, 1, 1 and 3. These may now be written as the denominators of a continued fraction as follows: A 1 C 2 + 1 1 + E 1 G B D 2 + F 1 3 H + J 1 1 K + L 1 1 M + N 1 1 + 1 3 We now have to find what are called the "convergents" of this fraction. In simpler terms, the convergents are approximations to the actual value of the fraction. They are, in value, alternately too large and too small, but each time they approach nearer to the actual value (hence the term convergent). 1st convergent [Down to line AB] = y 2nd convergent [To line CD] 3rd convergent [To line EF] 1 2+1 3 1 1 1 l 2+1 1 + 1 2+ 1 "2 + f 2 4th convergent [to line GH] 2+1 2+1 1 + I 1 + j_ 2 + J_ 2± 3 CONTINUED FRACTIONS 233 2+1 2+ I 2 + ^ ft K 7 _ 1 -10 — 27 1 +f ^ 5th convergent [to line JK] = 2 + \ + 1 2+1 1 + 1 1 + 1 2 + 1 2 + i 3 + 1 1 1 = 1 = 1 = I 2+1 2 + J_ 2 + & H u 9 1+1 u 6th convergent [to line LM] = y + \ 1 + 1 2+ 1 3 + 1 1 + 1 I 2+ 1 2 + 1 2 + 1 1 + 1 2 + 1 3 + i 1 + 1 2 + f 1 + 1 16 7 1 2 + 1 1 2 + 1 1 z 23 . 1 _23 • 62 ~ 62- 23 1 + 7 16 23 16 7th convergent [to line NP] . (The reader should now be able to check this for himself) = " 8th convergent [whole fraction] - m 353 The values of the convergents, their decimal equivalents and the error (+ or -) from the true value of the original fraction, are set down in the following table. 234 APPENDIX VII Original Fraction = Mi = 0-3711 Convergent No. Value. Decimal Error Equiv. + or -. 0-5000 + 0-1289 0-3333 -0-0378 0-3750 + 0-0039 0-3704 -0-0007 0-3714 +0-0003 0-3710 -0-0001 0-37113 -rG.00003 0-3711 0-0000 If a graph of the error against the convergent No. is plotted it will be somewhat like Fig. 159. It will be noticed that the error is only 0-0007 even at the 4th convergent, although this is a very simple fraction <#)• + V. O u \ 2 A^ 4 ^_ 6 1 1 / 3 Xi^ 5 Convergent number " 7 Fig. 159 Answers to exercises Exercises 2a Hole Shaft Clearance (mm) (mm) (mm) 1. 75-120 74-940 max 0-226 75-000 74-894 min 0-060 2. 35-062 35-018 max 0-060 35-000 35-002 min -0-018 (interference). 3. 20-021 19-980 max 0-062 20-000 19-959 min 0-020 4. 57-046 57-083 max -0-007 57-000 57-053 min -0-083 (interference). 5. Centre distance 124-98 mm; "Not Go" diameter 12-079 mm 6. 40-0062 mm diameter 7. Open out "Go" and "Not Go" ends by 0-052 mm 8. Same fit with blocks on top limit. Additional clearance of 0017mm with pair of blocks on bottom limits 9. Yes. Reduce "Go" diameter by 0-0373 mm and "Not Go" by 0-036 mm 10. Max. 0-271 mm: Min. 0- 170mm Exercises 2b 1-01 2. 104 3. 1005 4. 108 1-10 1-60 103 1-70 9-00 100 1-60 16-00 2000 700 5000 25-00 1-70 6. 106 7. 1-02 8. 1-05 24.00 1-30 1-50 1-40 50.00 13-00 7-00 17-00 100-00 2500 1-09 10.(a) 1005 (b) 105 11. "Go" end. "Not go" end. 1-90 1-02 1-90 1005 105 3-00 1-90 12-00 117 1-90 75-00 600 20-00 1-80 900 25-00 600 800 20-00 10-00 12. 1-08 400 1000 235 236 ANSWERS TO EXERCISES Exercises 2c 1. 0-30 mm 3. 10-31 m 4. One end 41-40 mm higher 5. 0° 6' too steep 6. BC = 39-25 mm; CD = 29- 15 mm AD = 35-90mm;>> = 13-10mm;x = 26-93 7. True angle = 37° 43'; error 0° 1' 8. 500-09 mm; 0- 108 mm 9. 75-79mm. 10. 1016mm 11. 14-86mm Exercises 2d 1.35° 18' 2. 22° 20' 3. 47° 4' 4. (i) 39°46'; (ii)a = 8,6 = 16; (iii) 25-01 5. 25° 50' 6. a = 46-23 mm, a = 30° 18' 7. A = 152- 19 mm; B = 57-81 mm 8. AB = 103-5mm; 29° 32'; 34° 12' 1. 40-546 mm; 25°2' Exercises 2e 2. ll-67mm 3. A = 77-508mm; B = 76-008mm 4. (25 - H) 2 = 225 - ( 102 D ) 2 . 1546mm 5. d = j; 26-67mm 6. 9° 12' Exercises 2f 1. (a) 20-53mm; (b) 36-43mm; (c)0-767in 2. Normal to axis, 58-9 mm diagonally across plugs, 59-26 mm 3. 10-09mm 4. 12-94mm 5. 15-4mm 6. 52-804mm 7. 14-16 mm 8. 64-05 mm Exercises 3a 28 39 54 1 75 104 145 202 280 250 180 130 1 94 68 49 35 25 877 1260 1 1820 2630 rev/min 6 4 1 3 7 mm Dia 4. 41 3 rev/min; 237mm, 119 mm, 60mm mm Dia 5. 5500 rev/min 32-28; 500 Exercises 3b 35-51 h 3.25-2m/min 4. Tu " g g C g arb rl .o .0 . - 27; 73 6. Rake 12° 10', Clearance 14° 50' 2-8 = —r- approx v - 20 7. 2-61 mm; 21° 8. 15° 1 10. Rake 19° 36'; Clearance 15° 36' 9. a = 7° 18'; /5 = 22° 13' Exercises 3c 1. 57°28' 2. 1 in 6-128; 3. 29° 4. (a) 155 mm; (6) 0-65 mm; (c) 7° 32' ANSWERS TO EXERCISES 237 5. 805mm; ±5' for 3°55' to 4°2' 6. 7-07mm; 247mm; 70° 38' 7. Angle 41° 18': Depth 6-95 mm Land 2-81 mm 8. 4-78 mm; l-66mm 9. 2-564 mm; 6122mm; 22° 03' with Horiz. 10. (a) 204mm; (b) 29-6 mm; (c) 11° 23' Exercises 3d 1. *J = jjp x 6 * ; i.5009mm 2. |J; Actual lead 3-1428 mm; Error + 0-0012mm 55 50 x 110 35 39 3. 4-433 mm; nearest 4-432 with -rj ratio; 66-47 mm/min 44 4. (a) 1-93 kW; (6)0-778mm 3 /J 5.312mm; 6.10mm 7. 384/7 8. 27-96 Nm (a)0-8786kW; (b) 2-238 J/mm 3 9. 1400 rev/min; 14-63Nm; 2-79 kW 10. 53-3 m; 1-95/? 20 40 x 25 11. Nearest Ratio = 49 = 35 ?() ; Pitch = 2-041 mm Exercises 4a 1. 6-69 mm; (a) 69-52 mm (6) 0-49 mm 2. (a) w = 9-435 mm; h = 61 mm; (6) w = 1000mm; A = 6-37 mm 3. Cutting 0-06 mm too shallow. 4. Cutting 0-17mm too deep; Tooth 0-12mm thin; Space 0-13mm wide. 5. (a) w = 1110mm, h = 5-98mm; (i) w = 1104mm, h = 5-95mm 6. 7-38 mm; (a) 25738mm (b) 178-48 mm 7. 19346 mm 8. (a) 23-63 mm; (b) 2345 mm 9. (a) 1845mm; (b) 140-6 mm; (c)86-7mm; (rf)78mm Exercises 4b 1. P.D. = 210mm; T.D. = 218mm; Depth = 9mm 2. w = 7-808 mm; h = 4-0768 mm 3. T = 40T; P.D. = 200mm; T.D. = 208mm; t = 24f; p.d. = 120mm; t.d. = 128mm; Depth = 9mm 4. 179-7 mm 5. 0-215° 6. (a) 221-35mm; (b) 229-35mm; (c) 9mm; (rf) 1910mm; (e) 63. 7. (a) 27; (b) P.D. 124-71 mm, T.D. 132-71 mm; (c) 678mm; (</) 42. 8. 2; P.D. = 70-71 mm; T.D. = 74-71 mm; lead = 222mm; 4-5 mm 9. Gear: 45T, 120mmP.D.; 125 mm T.D. ; Helix angle, 20° 22'; lead, 1016mm Pinion: 30T, 80mm P.D., 85mm T.D.; Helix angle, 20° 22', lead, 677mm Exercises 4c 1. T.D. = 46-36 mm; R.D. = 3263mm; A = 9° 2'. 2. 35T; P.D. 11140mm; T.D. 1 17-76 mm; Centre Dist 75-70 mm; Throat rad. 16-82mm; Whole dia. 124-71 mm; 28-22mm [Face angle 75°.] 3. 6; A = 32° 29' 8-62 mm 4. Wheel: 75T; P.D.238-73mm; throat dia. 24509 mm; rad. 2745 mm. Worm: 5start; P.D.61-27mm; T.D.67-63mm; A = 14°34' 238 ANSWERS TO EXERCISES 5. T = 24; P. angle, 45°; Cone dist., 4242mm; Add. angle, 3°22'; Ded. angle, 4°14'; Face, 48°22'; Root, 40°46'; Whole dia., 63-32 mm; Tip dist., 28-23 mm. 6. Wheel: 39T; D = 195mm; Angles: P = 56° 18'; Add., 2° 27'; Ded., 3°3'; Face, 58°45'; Root, 53° 15'; Whole dia., 200-55mm. Pinion: 26f; d = 130mm; Angles: P = 33° 42'; Add., 2° 27'; Ded., 3° 3'; Face, 36° 9'; Root, 30° 39'; Whole dia. 138-32mm; Cone dist., 11716mm. 7. 5-9 mm; 5° 58'. 8. Wheel: 24T; P.D. = 120mm; Angles: P = 73° 54'; Add.,4°34';Ded.,5°43';Wholedia, 122-77 mm; Cone dist., 62-45 mm. Pinion: 18T; P.D. = 90mm; Angles: P = 46°6'; Add., 4°34'; Ded., 5°43'; Whole dia., 96-93 mm. Exercises 5a 1. (a) 15; (6)6-944mm; (c)4-88mm 2. 25 to 30; 31-19mm. 3. 7°31'; negative. 4. 9°26'. 5. 82j° 6. 4-82mm. Exercises 5b 1. 70rev/min, 189mm/min 2. 60480mm 3 , 2-94 kW 3. 3mm. 4. 2-04 min 5. Spiral mill. 6. 1-8 kW, 2-5 min, 0-15p. Exercises 5c [in the answers to Exercises 5c and 5d, whole numbers refer to complete turns of the crank, numerators of fractions to holes, and denominators to hole circles. When the fraction is a simple one (e.g. \) it has been left in that form. Numerators of gear ratios are drivers, and denominators are driven gears.] 1. (a) 3i; (b) 2f; (c) Iff; (d) I A; (e) |? ; (J) f; (g) f?; (h) ® 2. (a) 3£; (b) 2tf; (c) ljf; (d) lfc (e) ft; CO ff; fe)§fc (h) j? 3. (a) If; (b) 2{f; (c) 3f; (d) 5{f; («) 7&. 4. (a) Iff; (b) 3&; (c) 3ff; (J) 8i; (<?) 15£ 5. If; 3 mm 6. (a) ^ + tf ; (6) ^ + if; (c) § - A; (<*)£-* 40C.c cN - C« 8. (a) 8 holes; 20 circle; Gears, fj} X ?f; plate to turn in same direction as crank. (b) 20 holes; 27 circle; Gears, ff x ft; plate turn same direction as crank, (c) 8 holes; 20 circle; Gears, ^; plate to turn opposite to crank, (d) 10 holes, 33 circle; Gears, If X ft; plate to turn same direction as crank. [Note. — Other solutions are possible to 8(a), (b), (c) and (d).] 9. 16°52' 10. 5ff;8f|; lift; 14ff Exercises 5d 1. (a) 1^; 10° 36' 26"; (b) 4f; 41° 24'; (c) 8^; 75° 45' 2. If. fl^ is nearer and could be obtained on Cincinnati.] , , 4 ... 100 x 32 ... 100 x 24 , , 32 40 x 24 3 - 7 * 4 - {a) :48x56 s (6) W4T ; (C) 72 ; (rf) 64-lT48 ANSWERS TO EXERCISES 239 5. 40X 32 ; a = 36°48' 48 x 56 6. ft X if; 24° 1' between axes of work and cutter 7. 720mm lead; tf ratio; 19° 14' 8. Nearest = ft X Jg (based on 245mm lead) 9. 72 mm; gear ratio f; a = 46°03' 10. Gear ratio f; a = 22° 1'; lead = 25 mm 11. Lead = 60 mm; gear ratio = f ; a = 30° UVofe. — Other solutions are possible to Nos. 9, 10 and 11.] Exercises 5e 1. 29°58' 2. 67°23' 3. 49°6' 4. (a) 41°6'; (b) 97°2' 5. 55-4 mm; 245-8 mm; 12°32' 6. (a) 50°46'; (6) 104°28' Exercises 6a [in the solutions to Ex. 6a the angle given is that made with the vertical or horizontal drawn to the commencing end of the vector.] 2. (a) 7-21 upwards; 34° to R of vert (b) 10-85 downwards to L; 41° below horiz (c) 10-94 downwards; 4° to R of vert (d) 31 R to L; 44° below horiz (e) 3-4 downwards; 13° to L of vert if) 12 L to R; 5° below horiz 3. (a) 7-2 upwards; 34° to L of vert (b) 10-85 upwards; 41° above horiz (c) 2-2 upwards; 21° to R of vert (d) 13-4 R to L; 32° above horiz 4. (<?) 14-5 upwards; 50° to R of vert (/) 9-3 downwards; 12° to L of vert 5. ac — 6-6; cb = 5-5; abc = 50°. 6. 5 units L to R; 37° above horiz 7. ac = 21; cZ> = 7-6 vert downwards. 8. ab = 8-7 L to R; 30° above horiz; be = 5 downwards; 30° to R of vert Exercises 6b 1. R to L (outwards); 11° 18' to lathe centre line. 2. Horiz 38-3N; vert 32-1 N 3. 212N; 53N/mm 2 4. 0-085m/s; 45° to horiz. 5. 266-3N 6. Suspension chain 196N; pulling chain 100N 7. 233N; 552N Exercises 6c 1. 1 530N in a plane at 34° to vertical; line of action inclined at 71° to 500N force. 2. 11mm 3. 309N; 904N 4. 216N.; 34° to horiz. 5. 130N; 200N 6. 520N; 26° to horiz. 7. 163-5N horiz. L. to R. 8. 2860N 9. 274mm rad; 100° from 50 kg mass 10. 5550N 11. 76-5cm 3 ; 113° from A 240 ANSWERS TO EXERCISES Exercises 6d 1. (a) 5-36 m/s,; (b) 4-64 m/s 2. 0-84 m/s 4. l-3m/s, 0-775mm 3 5. 0-17m/s; 23° to OA 7. 0-088 m/s 8. 856N 10. (a) f; (6) 0-244m/s. 11. 0-00052m/s 3. (a) 114; (b) 16-8 rpm. 6. 0-8 m/s 9. 0-0051 m/s; 7840 N Exercises 6e 1. 170N 2. 8N 3. Headstock 319N; tailstock 531N 4. Front bearing 1620N upwards; rear bearing 417N downwards. 5. Right 357N; left, 340N 6. 1600N, 600N 7. 177N 8. 80N 1. 4. 25-6N; 25-4N at 6° 57' 270N to horiz. Exercises 7a 2. 750N 5. 0-1184kW 3. 6. 12 300N 0-792 kW 1. 5. 9. 36-4; 93-9N 2. 68-5%; 18 060N 6. 2340J, 385-7N 10. l-97kW 121N 1200N Exercise 7b 3. 51-9N 7. 377; 5-3N 11. lONm Exercises 7c 4. 118N 8. 35-1% 1. d = 44-7mm; L = 894mm 2. RH load = 1080N; B. press = 0-27 N/mm 2 LH load = 1170N; B. press = 0-29 N/mm 2 3. n = 0-0158, 52-2 J 4. 86-2 mm 6. 50 300N 7. 7950N 63-2N 5. fi = 0032, 114kW 8. 50-9N/mm 2 ; 0-204mm 9. 23-1 N/mm 2 1. 10rev/min 2 ; 75 revs 4. 14-4 m/s 2 7. lm;5s 10. Stress = 200N/mm 2 ; tension = 40 000N Exercises 8a 2. 1-28 s, 12-5 m/s 3. 0-333 m/s 2 ; 1-2 s 5. 0-4m/s 2 ; 0-6m/s 6. 180rev/min 2 ; 1 min 8. 0-5m/s; 15s Exercises 8b 1. ION 2. lm/s 3. 355N 5. 17-3N 6. l-28m/s 2 ; 192N; 0-96m/s; 0-1849kW 7. 10-8m/s; 5890m/s 2 ; 589kN 8. 320mm 4. 72N 1. 18-31; 1050. 4. 157 500 J; 5250N 1. 22 700 J 5. 131N Exercises 8c 2. 6-55 s 6. 18-75J; 31-25N Exercises 8d 2. 6940 J 3. 101 N 6. 16-5rev/min 7. 6320N 3. 14-9rad/s 2 7. 8 J; 320 J 4. 16-5 s 8. 13-6rev/min ANSWERS TO EXERCISES 241 1. 1664 MJ 4. 785°C 1. 120kJ; 4-8°C 4. 9680J; 2-24°C Exercises 8e 2. 59°C 3. 12m 3 5. 38-61itres/min Exercises 8f 6. Cost gas 3-09 Cost coal — 1 2. 12m 3 /h 3. 264 000 J; 206kg Index Acceleration, 198 due to gravity, 200 Angles, cutting tool, 63 (solid), evaluation of, 142 Angular indexing, 130 measurement with sine bar, 22 surfaces, location of points on, 32 velocity and acceleration, 209 Arithmetical progression, 57 Backlash, 95 Balancing of masses in one plane, 159 Bar (pressure), 5 Bearing friction, 192 pressure, 191 Bevel gearing, 106 Bores, measuring large bores, 28 BSI Limit System, 14 Cam milling, 138 Celsius temperature, 218 Centigrade temperature, 218 Change wheels (odd threads), 77 Circular motion, 209 Coefficient of friction, 177 Continued fraction, 231 Conventions SI, 4 Cosecant, 92 Cotangent, 92 Cutters (milling), clearance, 117 fluting angle, 115 grinding, 117 number of teeth, 113 speeds and feeds, 121 tooth rake, 114 Cutting power, 79 speed range, 58 tool angles, 63 forces acting on, 157 life, 61 Dividing head, 123 Drilling power, 8 1 Efficiency of a machine, 181 Energy, 206 heat, 218, 222 of rotating bodies, 2 1 5 Equations, of motion, 198 Equilibrium, conditions for, 155, 171 Faceplate, balancing of work on, 159 Feeds for milling, 121 Fluting angle for milling cutter teeth, 115 Force, 203 of hammer blow, 205 Forces, acting on a cutting tool, 1 57 in a mechanism, 167 vectorial representation, 1 5 1 Form tools, 68 Fraction, continued, 231 Friction, 176 at a bearing 192 at nut face, 186 coefficient, 177 in screw thread, 182 of sliding keys, 188 when clamping, 178 Gauge, point for bores, 28 slip, 17 Gauging large radii, 25 Gear tooth form, 85 rack, 86 vernier, 87 Gearing, backlash, 95 base pitch, 92 bevel, 106 helical, 96 plug method of checking, 90 242 INDEX 243 stub, 94 worm, 102 Geometric speed range, 58 Graph of motion, 10, 199 Hammer blow, force of, 205 Heat, 218 of work, 222 specific capacity, 219 Helical gearing, 96 Helix, lead and angle, 98 Inclined plane, 182 Indexing, angular, 130 compound, 126 differential, 128 simple, 125 Involute, 84 ISO Limit system, 14 ISO Metric thread, 48 Johannson gauges, 17 Kelvin, 219 Kilogramme, 6 Level (spirit), 19 Limit, 13 systems, 14 Lines on angular surfaces, true length, 32 Mass, 6, 203 Measurement of gears, 87-90 by slip gauges, 17 of large bores, 28 of large radii, 25 of tapers, 23, 38, Mechanical advantage, 181 Metre, 2 Milling cam, 138 cutters, 113, feeds, 121 fluting angle, 115 indexing, 125 et seq power, 220 spiral, 134 Modulus of Elasticity, 195 Moment of a force, 170 Momentum, 203 Motion in a circle, 209 equations of, 198 Newton (force), 3 Pitch (base) of gears, 92 Power for cutting, 79, 122 Progression, arithmetic and geometric, 57 Rack (gear), 86 Radian, 209 Rake on milling cutter teeth, 1 14 Screw cutting, calculations for odd threads, 77 thread as inclined plane, 182 threads, measurement by wires, 47 Secant, 92 Sine bar, 22 SI units, 3 Slip gauges, 17 Specific heat capacity, 219 Speed range geometric, 58 Spiral milling, 134 Spirit level, 19 Strain, 193 Stress, 193 Stub teeth, 94 Taper, measurement with balls and rollers, 38 sine bar, 23 turning, 66 Temperature, 218 Tolerance, 13 Tool angles, 63 life calculations, 61 Tools, form, 68 Torque to cause acceleration, 211 Trigonometrical addition formulae, 230 Turning moment, 211 Units SI metric, 3 Vector diagrams of velocity, 162 244 INDEX Vectors, 148 addition and subtraction, 148 applications, 151 Velocity ratio of a machine, 181 vectorial representation, 162 Vernier (gear tooth) calculations, 87 Whitworth thread, 48 Wire measurement of screw threads, 47 Work, 206 Worm gearing, 102 Young's Modulus, 195 'This is a new edition with a new format, and completely revised to present its material in SI units; the introductory chapter leads straight into the "new" units and includes some worked exercises with them so the reader is left in no doubt as to what is to follow. 'The book divides neatly into two parts, the first dealing with machine tool calculations, liberally spaced in the text and very clearly illustrated. Each section is concluded by a number of problems to be solved by the student; answers are supplied at the end of the text. "The second part of the book deals with mechanic's principles, including statics, dynamics and a few pages on heat, invariably a product of a machine operation. The text concentrates its attention on the practical aspects of the principles involved and rightly leaves the more academic con- siderations to other authors and other places. 'The author and his books on workshop technology have enjoyed great popularity for 30 years, the books being standard works for many technical colleges: it is easy to see the reason for the appeal, for this edition is neat and compact yet the material is not over-compressed. We can expect to see it in use for another 30 years.' Marine Engineers Review o Other books in SI units by Dr Chapman: Elementary Workshop Calculations Workshop Technology Part 1, fifth edition Workshop Technology Part 2, fourth edition Workshop Technology Part 3, third edition EDWARD ARNOLD £2.50 net ISBN: 7131 3260 4